你可以制作一块电路板,用单片机控制的,这可以用C语言来写单片机的控制程序,然后将这块电路板连接到计算机串口或并口或usb口,用C写一个计算机程序,通过控制这块电路板实现对家用电器的控制,如果加上遥控模块,可以通过计算机遥控各种设备,想想你在外面玩电脑控制家里的电饭堡,微波炉吧,也可以用手机发消息来控制,这不是想象,是我已经做出的产品. 对了,就是C, 也有Delphi(Pascal)实现的版本.
算法大全（C，C++）一、 数论算法
1．求两数的最大公约数
function gcd(a,b:integer):integer
begin
if b=0 then gcd:=a
else gcd:=gcd (b,a mod b)
end
2．求两数的最小公倍数
function lcm(a,b:integer):integer
begin
if a<b then swap(a,b)
lcm:=a
while lcm mod b>0 do inc(lcm,a)
end
3．素数的求法
A.小范围内判断一个数是否为质数：
function prime (n: integer): Boolean
var I: integer
begin
for I:=2 to trunc(sqrt(n)) do
if n mod I=0 then begin
prime:=falseexit
end
prime:=true
end
B.判断longint范围内的数是否为素数（包含求50000以内的素数表）：
procedure getprime
var
i,j:longint
p:array[1..50000] of boolean
begin
fillchar(p,sizeof(p),true)
p[1]:=false
i:=2
while i<50000 do begin
if p[i] then begin
j:=i*2
while j<50000 do begin
p[j]:=false
inc(j,i)
end
end
inc(i)
end
l:=0
for i:=1 to 50000 do
if p[i] then begin
inc(l)pr[l]:=i
end
end{getprime}
function prime(x:longint):integer
var i:integer
begin
prime:=false
for i:=1 to l do
if pr[i]>=x then break
else if x mod pr[i]=0 then exit
prime:=true
end{prime}
二、图论算法
1．最小生成树
A.Prim算法：
procedure prim(v0:integer)
var
lowcost,closest:array[1..maxn] of integer
i,j,k,min:integer
begin
for i:=1 to n do begin
lowcost[i]:=cost[v0,i]
closest[i]:=v0
end
for i:=1 to n-1 do begin
{寻找离生成树最近的未加入顶点k}
min:=maxlongint
for j:=1 to n do
if (lowcost[j]<min) and (lowcost[j]<>0) then begin
min:=lowcost[j]
k:=j
end
lowcost[k]:=0{将顶点k加入生成树}
{生成树中增加一条新的边k到closest[k]}
{修正各点的lowcost和closest值}
for j:=1 to n do
if cost[k,j]<lwocost[j] then begin
lowcost[j]:=cost[k,j]
closest[j]:=k
end
end
end{prim}
B.Kruskal算法：(贪心)
按权值递增顺序删去图中的边，若不形成回路则将此边加入最小生成树。
function find(v:integer):integer{返回顶点v所在的集合}
var i:integer
begin
i:=1
while (i<=n) and (not v in vset[i]) do inc(i)
if i<=n then find:=i else find:=0
end
procedure kruskal
var
tot,i,j:integer
begin
for i:=1 to n do vset[i]:=[i]{初始化定义n个集合，第I个集合包含一个元素I}
p:=n-1q:=1tot:=0{p为尚待加入的边数，q为边集指针}
sort
{对所有边按权值递增排序，存于e[I]中，e[I].v1与e[I].v2为边I所连接的两个顶点的序号，e[I].len为第I条边的长度}
while p>0 do begin
i:=find(e[q].v1)j:=find(e[q].v2)
if i<>j then begin
inc(tot,e[q].len)
vset[i]:=vset[i]+vset[j]vset[j]:=[]
dec(p)
end
inc(q)
end
writeln(tot)
end
A.标号法求解单源点最短路径：
var
a:array[1..maxn,1..maxn] of integer
b:array[1..maxn] of integer{b[i]指顶点i到源点的最短路径}
mark:array[1..maxn] of boolean
procedure bhf
var
best,best_j:integer
begin
fillchar(mark,sizeof(mark),false)
mark[1]:=trueb[1]:=0{1为源点}
repeat
best:=0
for i:=1 to n do
If mark[i] then {对每一个已计算出最短路径的点}
for j:=1 to n do
if (not mark[j]) and (a[i,j]>0) then
if (best=0) or (b[i]+a[i,j]<best) then begin
best:=b[i]+a[i,j]best_j:=j
end
if best>0 then begin
b[best_j]:=best；mark[best_j]:=true
end
until best=0
end{bhf}
B.Floyed算法求解所有顶点对之间的最短路径：
procedure floyed
begin
for I:=1 to n do
for j:=1 to n do
if a[I,j]>0 then p[I,j]:=I else p[I,j]:=0{p[I,j]表示I到j的最短路径上j的前驱结点}
for k:=1 to n do {枚举中间结点}
for i:=1 to n do
for j:=1 to n do
if a[i,k]+a[j,k]<a[i,j] then begin
a[i,j]:=a[i,k]+a[k,j]
p[I,j]:=p[k,j]
end
end
C. Dijkstra 算法：
var
a:array[1..maxn,1..maxn] of integer
b,pre:array[1..maxn] of integer{pre[i]指最短路径上I的前驱结点}
mark:array[1..maxn] of boolean
procedure dijkstra(v0:integer)
begin
fillchar(mark,sizeof(mark),false)
for i:=1 to n do begin
d[i]:=a[v0,i]
if d[i]<>0 then pre[i]:=v0 else pre[i]:=0
end
mark[v0]:=true
repeat {每循环一次加入一个离1集合最近的结点并调整其他结点的参数}
min:=maxintu:=0{u记录离1集合最近的结点}
for i:=1 to n do
if (not mark[i]) and (d[i]<min) then begin
u:=imin:=d[i]
end
if u<>0 then begin
mark[u]:=true
for i:=1 to n do
if (not mark[i]) and (a[u,i]+d[u]<d[i]) then begin
d[i]:=a[u,i]+d[u]
pre[i]:=u
end
end
until u=0
end
3.计算图的传递闭包
Procedure Longlink
Var
T:array[1..maxn,1..maxn] of boolean
Begin
Fillchar(t,sizeof(t),false)
For k:=1 to n do
For I:=1 to n do
For j:=1 to n do T[I,j]:=t[I,j] or (t[I,k] and t[k,j])
End
4．无向图的连通分量
A.深度优先
procedure dfs ( now,color: integer)
begin
for i:=1 to n do
if a[now,i] and c[i]=0 then begin {对结点I染色}
c[i]:=color
dfs(I,color)
end
end
B 宽度优先（种子染色法）
5．关键路径
几个定义： 顶点1为源点，n为汇点。
a. 顶点事件最早发生时间Ve[j], Ve [j] = max{ Ve [j] + w[I,j] },其中Ve (1) = 0
b. 顶点事件最晚发生时间 Vl[j], Vl [j] = min{ Vl[j] – w[I,j] },其中 Vl(n) = Ve(n)
c. 边活动最早开始时间 Ee[I], 若边I由<j,k>表示，则Ee[I] = Ve[j]
d. 边活动最晚开始时间 El[I], 若边I由<j,k>表示，则El[I] = Vl[k] – w[j,k]
若 Ee[j] = El[j] ，则活动j为关键活动，由关键活动组成的路径为关键路径。
求解方法：
a. 从源点起topsort,判断是否有回路并计算Ve
b. 从汇点起topsort,求Vl
c. 算Ee 和 El
6．拓扑排序
找入度为0的点，删去与其相连的所有边，不断重复这一过程。
例 寻找一数列，其中任意连续p项之和为正，任意q 项之和为负，若不存在则输出NO.
7.回路问题
Euler回路(DFS)
定义：经过图的每条边仅一次的回路。（充要条件：图连同且无奇点）
Hamilton回路
定义：经过图的每个顶点仅一次的回路。
一笔画
充要条件：图连通且奇点个数为0个或2个。
9．判断图中是否有负权回路 Bellman-ford 算法
x[I],y[I],t[I]分别表示第I条边的起点，终点和权。共n个结点和m条边。
procedure bellman-ford
begin
for I:=0 to n-1 do d[I]:=+infinitive
d[0]:=0
for I:=1 to n-1 do
for j:=1 to m do {枚举每一条边}
if d[x[j]]+t[j]<d[y[j]] then d[y[j]]:=d[x[j]]+t[j]
for I:=1 to m do
if d[x[j]]+t[j]<d[y[j]] then return false else return true
end
10．第n最短路径问题
*第二最短路径：每举最短路径上的每条边，每次删除一条，然后求新图的最短路径，取这些路径中最短的一条即为第二最短路径。
*同理，第n最短路径可在求解第n-1最短路径的基础上求解。
三、背包问题
*部分背包问题可有贪心法求解：计算Pi/Wi
数据结构：
w[i]:第i个背包的重量；
p[i]:第i个背包的价值；
1．0-1背包： 每个背包只能使用一次或有限次(可转化为一次)：
A.求最多可放入的重量。
NOIP2001 装箱问题
有一个箱子容量为v(正整数，o≤v≤20000)，同时有n个物品(o≤n≤30)，每个物品有一个体积 (正整数)。要求从 n 个物品中，任取若千个装入箱内，使箱子的剩余空间为最小。
l 搜索方法
procedure search(k,v:integer){搜索第k个物品，剩余空间为v}
var i,j:integer
begin
if v<best then best:=v
if v-(s[n]-s[k-1])>=best then exit{s[n]为前n个物品的重量和}
if k<=n then begin
if v>w[k] then search(k+1,v-w[k])
search(k+1,v)
end
end
l DP
F[I,j]为前i个物品中选择若干个放入使其体积正好为j的标志，为布尔型。
实现:将最优化问题转化为判定性问题
f [I, j] = f [ i-1, j-w[i] ] (w[I]<=j<=v) 边界：f[0,0]:=true.
For I:=1 to n do
For j:=w[I] to v do F[I,j]:=f[I-1,j-w[I]]
优化：当前状态只与前一阶段状态有关，可降至一维。
F[0]:=true
For I:=1 to n do begin
F1:=f
For j:=w[I] to v do
If f[j-w[I]] then f1[j]:=true
F:=f1
End
B.求可以放入的最大价值。
F[I,j] 为容量为I时取前j个背包所能获得的最大价值。
F [i,j] = max { f [ i – w [ j ], j-1] + p [ j ], f[ i,j-1] }
C.求恰好装满的情况数。
DP:
Procedure update
var j,k:integer
begin
c:=a
for j:=0 to n do
if a[j]>0 then
if j+now<=n then inc(c[j+now],a[j])
a:=c
end
2．可重复背包
A求最多可放入的重量。
F[I,j]为前i个物品中选择若干个放入使其体积正好为j的标志，为布尔型。
状态转移方程为
f[I,j] = f [ I-1, j – w[I]*k ] (k=1.. j div w[I])
B.求可以放入的最大价值。
USACO 1.2 Score Inflation
进行一次竞赛，总时间T固定，有若干种可选择的题目，每种题目可选入的数量不限，每种题目有一个ti（解答此题所需的时间）和一个si（解答此题所得的分数），现要选择若干题目，使解这些题的总时间在T以内的前提下，所得的总分最大，求最大的得分。
*易想到：
f[i,j] = max { f [i- k*w[j], j-1] + k*p[j] } (0<=k<= i div w[j])
其中f[i,j]表示容量为i时取前j种背包所能达到的最大值。
*实现：
Begin
FillChar(f,SizeOf(f),0)
For i:=1 To M Do
For j:=1 To N Do
If i-problem[j].time>=0 Then
Begin
t:=problem[j].point+f[i-problem[j].time]
If t>f[i] Then f[i]:=t
End
Writeln(f[M])
End.
C.求恰好装满的情况数。
Ahoi2001 Problem2
求自然数n本质不同的质数和的表达式的数目。
思路一，生成每个质数的系数的排列，在一一测试，这是通法。
procedure try(dep:integer)
var i,j:integer
begin
cal{此过程计算当前系数的计算结果，now为结果}
if now>n then exit{剪枝}
if dep=l+1 then begin {生成所有系数}
cal
if now=n then inc(tot)
exit
end
for i:=0 to n div pr[dep] do begin
xs[dep]:=i
try(dep+1)
xs[dep]:=0
end
end
思路二，递归搜索效率较高
procedure try(dep,rest:integer)
var i,j,x:integer
begin
if (rest<=0) or (dep=l+1) then begin
if rest=0 then inc(tot)
exit
end
for i:=0 to rest div pr[dep] do
try(dep+1,rest-pr[dep]*i)
end
{main: try(1,n)}
思路三：可使用动态规划求解
USACO1.2 money system
V个物品，背包容量为n，求放法总数。
转移方程：
Procedure update
var j,k:integer
begin
c:=a
for j:=0 to n do
if a[j]>0 then
for k:=1 to n div now do
if j+now*k<=n then inc(c[j+now*k],a[j])
a:=c
end
{main}
begin
read(now){读入第一个物品的重量}
i:=0{a[i]为背包容量为i时的放法总数}
while i<=n do begin
a[i]:=1inc(i,now)end{定义第一个物品重的整数倍的重量a值为1，作为初值}
for i:=2 to v do
begin
read(now)
update{动态更新}
end
writeln(a[n])
四、排序算法
A.快速排序：
procedure qsort(l,r:integer)
var i,j,mid:integer
begin
i:=lj:=rmid:=a[(l+r) div 2]{将当前序列在中间位置的数定义为中间数}
repeat
while a[i]<mid do inc(i){在左半部分寻找比中间数大的数}
while a[j]>mid do dec(j){在右半部分寻找比中间数小的数}
if i<=j then begin {若找到一组与排序目标不一致的数对则交换它们}
swap(a[i],a[j])
inc(i)dec(j){继续找}
end
until i>j
if l<j then qsort(l,j){若未到两个数的边界，则递归搜索左右区间}
if i<r then qsort(i,r)
end{sort}
B.插入排序：
思路：当前a[1]..a[i-1]已排好序了，现要插入a[i]使a[1]..a[i]有序。
procedure insert_sort
var i,j:integer
begin
for i:=2 to n do begin
a[0]:=a[i]
j:=i-1
while a[0]<a[j] do begin
a[j+1]:=a[j]
j:=j-1
end
a[j+1]:=a[0]
end
end{inset_sort}
C.选择排序：
procedure sort
var i,j,k:integer
begin
for i:=1 to n-1 do
for j:=i+1 to n do
if a[i]>a[j] then swap(a[i],a[j])
end
D. 冒泡排序
procedure bubble_sort
var i,j,k:integer
begin
for i:=1 to n-1 do
for j:=n downto i+1 do
if a[j]<a[j-1] then swap( a[j],a[j-1]){每次比较相邻元素的关系}
end
E.堆排序：
procedure sift(i,m:integer){调整以i为根的子树成为堆,m为结点总数}
var k:integer
begin
a[0]:=a[i]k:=2*i{在完全二叉树中结点i的左孩子为2*i,右孩子为2*i+1}
while k<=m do begin
if (k<m) and (a[k]<a[k+1]) then inc(k){找出a[k]与a[k+1]中较大值}
if a[0]<a[k] then begin a[i]:=a[k]i:=kk:=2*iend
else k:=m+1
end
a[i]:=a[0]{将根放在合适的位置}
end
procedure heapsort
var
j:integer
begin
for j:=n div 2 downto 1 do sift(j,n)
for j:=n downto 2 do begin
swap(a[1],a[j])
sift(1,j-1)
end
无聊，粘着玩 657行#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <stdarg.h>
#include <ctype.h>
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <conio.h>
#include <Winsock2.h>
#include <ws2tcpip.h>
#include <io.h>
int raw2iframe(char *rawfile, char *ifile )
int Ts_pes(const char *src, const char *des ,unsigned short pid)
int pes_raw( char *pesfile, char *rawfile )
int es_pes(char *src, char *des)
int pes_ts(char *tsfile, char *pesfile)
static unsigned char m_buf[188 * 1024]
static unsigned char buf[188*1024]
/* 设置Dts时间戳(90khz) */
unsigned int SetDtsTimeStamp( unsigned char *buf, unsigned int time_stemp)
{
buf[0] = ((time_stemp >>29) | 0x11 ) &0x1f
buf[1] = time_stemp >>22
buf[2] = (time_stemp >>14) | 0x01
buf[3] = time_stemp >>7
buf[4] = (time_stemp <<1) | 0x01
return 0
}
/* 设置Pts时间戳(90khz) */
unsigned int SetPtsTimeStamp( unsigned char *buf, unsigned int time_stemp)
{
buf[0] = ((time_stemp >>29) | 0x31 ) &0x3f
buf[1] = time_stemp >>22
buf[2] = (time_stemp >>14) | 0x01
buf[3] = time_stemp >>7
buf[4] = (time_stemp <<1) | 0x01
return 0
}
/* 读取时戳(45khz) */
unsigned int GetTimeStamp( unsigned char *buf )
{
unsigned int ts
unsigned char *p = buf
ts = ((*p>>1) &0x7) <<30
p++
ts += (*p) <<22
p++
ts += ((*p)>>1) <<15
p++
ts += (*p) <<7
p++
ts += (*p) >>1
p++
return ts
}
/*主函数*/
int main(void)
{
Ts_pes("shoes.ts", "641.pes", 641) //提取PES
pes_raw("641.pes","641.raw") //提取ES
raw2iframe( "641.raw", "641.iframe" )//提取I帧
es_pes("641.iframe", "es.pes") //打包成PES
pes_ts("pes.ts","es.pes")//打包成TS
return 0
}
/*从视频中提取PES*/
int Ts_pes(const char *src, const char *des ,unsigned short pid)
{
unsigned char *p
FILE *fpr, *fpw
int size
int ret
unsigned short vpid
unsigned char adaptation_field_control
unsigned char continuity_counter
unsigned char adaptation_field_length
unsigned char *payload
unsigned char payload_unit_start_indicator
unsigned char last_counter = 0xff
int start = 0
fpr = fopen(src, "rb") //打开文件
fpw = fopen(des, "wb") //写入文件
if(NULL == fpr || NULL == fpw)
{
return -1
}
size = sizeof(m_buf)
while(!feof(fpr))
{
ret = fread(m_buf, 1, size, fpr )//读取文件
p = m_buf
while(p <m_buf + ret)
{
vpid = (p[1] &0x1f) <<8 | p[2]
if(pid == vpid)
{
adaptation_field_control = (p[3]>>4)&0x3//判断是否有可调整字段
continuity_counter = p[3] &0xf
payload_unit_start_indicator = (p[1]>>6) &0x1
payload = NULL
adaptation_field_length = p[4]
switch(adaptation_field_control)
{
case 0:
case 2:
break /*0为保留，2为只有调整无负载*/
case 1:
payload = p + 4/*无调整字段*/
break
case 3:
payload = p + 5 + p[4]/*净荷*/
break
}
if(1 == payload_unit_start_indicator)
{
start = 1
}
if(start &&payload)
{
fwrite(payload, 1, p + 188 - payload, fpw) //写入文件
}
if( last_counter!= 0xff &&((last_counter +1 )&0xf) != continuity_counter )
{
printf( "data lost\n" )
}
last_counter = continuity_counter
}
p = p + 188
}
}
printf("ts_pes_END\n")
fclose(fpr) //关闭
fclose(fpw)
return 0
}
/*从PES中提取ES*/
int pes_raw( char *pesfile, char *rawfile )
{
FILE *fpd, *fp
unsigned char *p, *payload, *tmp
int size, num, rdsize
unsigned int last = 0
__int64 total = 0, wrsize = 0
fp = fopen( pesfile, "rb" )
fpd = fopen( rawfile, "wb" )
if( fp == NULL || fpd == NULL )
return -1
num = 0
size = 0
p = m_buf
while( true )
{
REDO:
if( m_buf + size <= p )
{
p = m_buf
size = 0
}
else if( m_buf <p &&p <m_buf + size )
{
size -= p - m_buf
memmove( m_buf, p, size )
p = m_buf
}
if( !feof(fp) &&size <sizeof(m_buf) )
{
rdsize = fread( m_buf+size, 1, sizeof(m_buf)-size, fp )
size += rdsize
total += rdsize
}
if( size <= 0 )
break
tmp = p
/* 寻找PES-HEADER: 0X000001E0 */
while( p[0] != 0 || p[1] != 0 || p[2] != 0x01 ||
( ( p[3] &0xe0 ) != 0xe0 &&( p[3] &0xe0 ) != 0xc0 ) )
{
p++
if( m_buf + size <= p )
goto REDO
}
if( p != tmp )
{
printf( "pes skip size=%d\n", p - tmp )
}
/* PES_packet_length */
unsigned short len = (p[4]<<8) | p[5]
if( len == 0 )
{
unsigned char *end = p + 6
while( end[0] != 0 || end[1] != 0 || end[2] != 0x01 ||
( ( end[3] &0xe0 ) != 0xe0 &&( end[3] &0xc0 ) != 0xc0 ) )
{
if( m_buf + size <= end )
{
if( feof(fp) )
break
goto REDO
}
end++
}
len = end - p - 6
}
if( m_buf + size <p + 6 + len )
{
if( feof(fp) )
break
continue
}
p += 6
{
unsigned char PES_scrambling_control = (*p>>4)&0x3
unsigned char PES_priority = (*p>>3)&0x1
unsigned char data_alignment_indicator = (*p>>2)&0x1
unsigned char copyright = (*p>>1)&0x1
unsigned char original_or_copy = (*p)&0x1
p++
unsigned char PTS_DTS_flags = (*p>>6)&0x3
unsigned char ESCR_flag = (*p>>5)&0x1
unsigned char ES_rate_flag = (*p>>4)&0x1
unsigned char DSM_trick_mode_flag = (*p>>3)&0x1
unsigned char additional_copy_info_flag = (*p>>2)&0x1
unsigned char PES_CRC_flag = (*p>>1)&0x1
unsigned char PES_extension_flag = (*p)&0x1
p++
unsigned char PES_header_data_length = *p
p++
payload = p + PES_header_data_length
if (PTS_DTS_flags == 0x2 )
{
unsigned int pts
pts = (*p>>1) &0x7
pts = pts <<30
p++
pts += (*p)<<22
p++
pts += ((*p)>>1)<<15
p++
pts += (*p)<<7
p++
pts += (*p)>>1
p++
p -= 5
if( pts <last )
{
printf( "?\n" )
}
last = pts
}
else if( PTS_DTS_flags == 0x3 )
{
unsigned int pts, dts
pts = (*p>>1) &0x7
pts = pts <<30
p++
pts += (*p)<<22
p++
pts += ((*p)>>1)<<15
p++
pts += (*p)<<7
p++
pts += (*p)>>1
p++
dts = (*p>>1) &0x7
dts = dts <<30
p++
dts += (*p)<<22
p++
dts += ((*p)>>1)<<15
p++
dts += (*p)<<7
p++
dts += (*p)>>1
p++
p -= 10
printf( "num=%d dura=%d size=%d pts-dts=%d\n", num, pts - last, len-3-PES_header_data_length, (int)(pts-dts) )
if( pts <last )
{
printf( "?\n" )
}
last = pts
}
else if( PTS_DTS_flags != 0 )
{
printf( "error\n" )
}
if( fpd )
{
fwrite( p + PES_header_data_length, 1, len - 3 - PES_header_data_length, fpd )
wrsize += len - 3 - PES_header_data_length
}
num++
p += len - 3
}
payload = p
size -= p - m_buf
memmove( m_buf, p, size )
p = m_buf
}
fclose( fp )
fclose( fpd )
printf("pes_raw_END\n")
return 0
}
/*提取I帧*/
int raw2iframe(char *rawfile, char *ifile )
{
unsigned char *temp_p
unsigned char *p
unsigned char picture_coding_type
unsigned char buf[188*1024] = {0}
unsigned char pes_buf[32*1024] = {0}
unsigned short pid = 641
unsigned short t_pid = 0
int i = 0
int j = 0
int size = 0
int iLen = 0
int wiLen = 0
int temp_queue = 0
int temp_ifrem = 0
void *fps = CreateFile(rawfile, GENERIC_READ, FILE_SHARE_READ, NULL, OPEN_EXISTING, 0, NULL) //打开读文件
void *fpd = CreateFile(ifile, GENERIC_READ|GENERIC_WRITE, FILE_SHARE_READ, NULL, CREATE_ALWAYS, 0 , NULL) //打开写文件
temp_p = NULL
while(1)
{
if (0 == ReadFile(fps, buf+size, sizeof(buf)-size, (unsigned long *)&iLen, NULL)) //读取文件内容
{
CloseHandle((HANDLE)fps)
return -1
}
if (0 == iLen)
{
break
}
p = buf
while( p + 6 <buf + iLen +size)
{
if (p[0] == 0x0 &&p[1] == 0x0 &&p[2] == 0x1)//是头进入
{
if (p[3] == 0x0 || p[3] == 0xB3 )
{
if ( (NULL != temp_p) &&((1 == temp_queue) || (1 == temp_ifrem)))
{
WriteFile(fpd, temp_p, p-temp_p, (unsigned long *)&wiLen, NULL) //写文件，先写序列头，再写I帧
temp_queue = 0
temp_ifrem = 0
temp_p = NULL
}
}
if (p[3] == 0xB3) //判断是视频序列，则初始化
{
temp_queue = 1
temp_p = p
}
picture_coding_type = (p[5]>>3) &0x7
if (p[3] == 0x0 &&1 == picture_coding_type) //判断是I帧，则初始化
{
temp_ifrem = 1
temp_p = p
}
}
p++
}
/*把多出来的内容写入下个BUF*/
if(NULL != temp_p)
{
size = buf + iLen + size - temp_p
memmove(buf, temp_p, size)
}
else
{
size = buf + iLen + size - p
memmove(buf, p, size)
}
temp_p = NULL
}
CloseHandle((HANDLE)fps)
CloseHandle((HANDLE)fpd)
printf("raw_iframe_END\n")
return 0
}
/*打包成PES*/
int es_pes(char *src, char *des)
{
FILE *iframe_fp, *pes_fp
unsigned char *p
unsigned char *temp_p = NULL
unsigned char buf[188*1024] = {0}
unsigned char pes_header[19]
unsigned short int pes_packet_length = 0
unsigned int framecnt = 0
unsigned char flag = 0
unsigned int Pts = 0
unsigned int Dts = 0
int i = 0
int size = 0
int iLen = 0
int wiLen = 0
int temp_que = 0
iframe_fp = fopen( src, "rb" )
pes_fp = fopen( des, "wb" )
if( iframe_fp == NULL || pes_fp == NULL )
{
return -1
}
while(!feof(iframe_fp))
{
iLen = fread(buf + size, 1, sizeof(buf) - size, iframe_fp)
p = buf
while( p + 3 <buf + iLen +size)
{
memset(pes_header, 0, sizeof(pes_header))
if (p[0] == 0x0 &&p[1] == 0x0 &&p[2] == 0x1 &&0xB3 == p[3])
{
if ((NULL != temp_p) &&(1 == temp_que))
{
LAST_I:
pes_packet_length = p - temp_p + 13
pes_header[4] = (pes_packet_length&0xff00) >>8
pes_header[5] = pes_packet_length&0x00ff
/*PES包头的相关数据*/
pes_header[0] = 0
pes_header[1] = 0
pes_header[2] = 0x01
pes_header[3] = 0xE0
pes_header[6] = 0x81
pes_header[7] = 0xC0
pes_header[8] = 0x0A
Dts = (framecnt + 1) * 40 * 90
Pts = framecnt * 40 * 90
SetPtsTimeStamp(&pes_header[9], Pts) //设置时间戳 PTS
SetDtsTimeStamp(&pes_header[14], Dts)//设置时间戳 DTS
framecnt++
if (0 == flag)
{
fwrite(pes_header, 1, sizeof(pes_header), pes_fp) //把PES包头写入文件
fwrite(temp_p, 1, p-temp_p, pes_fp) //把I帧写入文件
}
else
{
fwrite(pes_header, 1, sizeof(pes_header), pes_fp) //把PES包头写入文件
fwrite(temp_p, 1, p-temp_p+3, pes_fp) //把I帧写入文件
}
temp_p = NULL
}
if (p[3] == 0xB3)//判断是否到了下一个序列头
{
temp_p = p
temp_que = 1
}
}
p++
}
/*把多出来的内容写入下个BUF*/
if(NULL != temp_p)
{
if (feof(iframe_fp)) //把最后一帧写入文件
{
//flag = 1
goto LAST_I
}
size = buf + iLen + size - temp_p
memmove(buf, temp_p, size)
}
else
{
size = buf + iLen + size - p
memmove(buf, p, size)
}
temp_p = NULL
}
printf("es_pes_END\n")
fclose(iframe_fp)
fclose(pes_fp)
return 0
}
/*打包成TS包*/
int pes_ts(char *tsfile, char *pesfile)
{
FILE *ts_fp, *pes_fp
int flag = 0
int iLen = 0
int size = 0
int temp_pes = 0
int pes_packet_len = 0
unsigned char *p
unsigned char counter = 0
unsigned char *temp_p = NULL
unsigned char ts_buf[188] = {0}
unsigned char start_indicator_flag = 0
pes_fp = fopen(pesfile, "rb")
ts_fp = fopen(tsfile, "wb")
if( ts_fp == NULL || pes_fp == NULL )
{
return -1
}
/*设ts参数*/
ts_buf[0] = 0x47
ts_buf[1] = 0x62
ts_buf[2] = 0x81
while(!feof(pes_fp))
{
iLen = fread(buf+size, 1, sizeof(buf)-size, pes_fp) //读文件
p = buf
while( p + 6 <buf + iLen +size)
{
if (0 == p[0] &&0 == p[1] &&0x01 == p[2] &&0xE0 == p[3]) //进入
{
if (flag == 0) //第一次找到PES包
{
temp_p = p
flag = 1
}
else
{
pes_packet_len = p - temp_p //pes包长度
start_indicator_flag = 0
while (1)
{
ts_buf[3] = counter
if (1 != start_indicator_flag)
{
ts_buf[1] = ts_buf[1] | 0x40 //payload_unit_start_indicator置为1
start_indicator_flag = 1
}
else
{
ts_buf[1] = ts_buf[1] &0xBF //payload_unit_start_indicator置为0
}
if (pes_packet_len >184) //打包成TS包（188）
{
ts_buf[3] = ts_buf[3] &0xDF
ts_buf[3] = ts_buf[3] | 0x10
memcpy(&ts_buf[4], temp_p, 184)
fwrite(ts_buf, 1, 188, ts_fp) //写文件
pes_packet_len -=184
temp_p += 184
}
else //不够184B的加入调整字段，为空
{
ts_buf[3] = ts_buf[3] | 0x30
ts_buf[4] = 183 - pes_packet_len
memcpy(&ts_buf[4] + 1 + ts_buf[4], temp_p, pes_packet_len)
fwrite(ts_buf, 1, 188, ts_fp) //写文件
break
}
counter = (counter + 1) % 0x10 //ts包计数
}
}
temp_p = p
}
p++
}
if (1 == flag)
{
size = buf + iLen + size - temp_p
memmove(buf, temp_p, size)
temp_p = NULL
flag = 0
}
else
{
size =buf + iLen + size - p
memmove(m_buf , p , size)
}
}
printf("pes_ts_END\n")
fclose(pes_fp)
fclose(ts_fp)
return 0
}
