本文概述了AI音视频处理领域的关键技术，包括视频分割、语义识别、自动驾驶、AR增强现实、SLAM、物体检测与识别、语音识别与变声等。探讨了这些技术在实际应用中的角色与价值。
题目：
Task Schedule
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5748 Accepted Submission(s): 1848
Problem Description
Our geometry princess XMM has stoped her study in computational geometry to concentrate on her newly opened factory. Her factory has introduced M new machines in order to process the coming N tasks. For the i-th task, the factory has to start processing it at or after day Si, process it for Pi days, and finish the task before or at day Ei. A machine can only work on one task at a time, and each task can be processed by at most one machine at a time. However, a task can be interrupted and processed on different machines on different days.
Now she wonders whether he has a feasible schedule to finish all the tasks in time. She turns to you for help.
Now she wonders whether he has a feasible schedule to finish all the tasks in time. She turns to you for help.
Input
On the first line comes an integer T(T<=20), indicating the number of test cases.
You are given two integer N(N<=500) and M(M<=200) on the first line of each test case. Then on each of next N lines are three integers Pi, Si and Ei (1<=Pi, Si, Ei<=500), which have the meaning described in the description. It is guaranteed that in a feasible schedule every task that can be finished will be done before or at its end day.
You are given two integer N(N<=500) and M(M<=200) on the first line of each test case. Then on each of next N lines are three integers Pi, Si and Ei (1<=Pi, Si, Ei<=500), which have the meaning described in the description. It is guaranteed that in a feasible schedule every task that can be finished will be done before or at its end day.
Output
For each test case, print “Case x: ” first, where x is the case number. If there exists a feasible schedule to finish all the tasks, print “Yes”, otherwise print “No”.
Print a blank line after each test case.
Print a blank line after each test case.
Sample Input
2 4 3 1 3 5 1 1 4 2 3 7 3 5 9 2 2 2 1 3 1 2 2
Sample Output
Case 1: Yes Case 2: Yes
Author
allenlowesy
Source
Recommend
zhouzeyong
题意：有N个任务，M台机器，任务i在Si开始，Ei之前结束，要用Pi天，每台机器每天只能执行一个任务，一个任务每天也只能交给一台机器执行，但可以在不同的天由不同的机器执行，问是否有安排方案满足。
思路：每个任务与源点建边，容量为Pi，与Si到Ei的每条边都连边，容量为1，每天与汇点连边，容量为M。
代码：
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include<climits>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;
#define PB push_back
#define MP make_pair
#define REP(i,x,n) for(int i=x;i<(n);++i)
#define FOR(i,l,h) for(int i=(l);i<=(h);++i)
#define FORD(i,h,l) for(int i=(h);i>=(l);--i)
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define OI(X) printf("%d",X);
#define RS(X) scanf("%s", (X))
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
#define Swap(a, b) (a ^= b, b ^= a, a ^= b)
#define Dpoint strcut node{int x,y}
#define cmpd int cmp(const int &a,const int &b){return a>b;}
/*#ifdef HOME
freopen("in.txt","r",stdin);
#endif*/
const int MOD = 1e9+7;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long LL;
typedef pair<int,int> PII;
//#define HOME
int Scan()
{
int res = 0, ch, flag = 0;
if((ch = getchar()) == '-') //判断正负
flag = 1;
else if(ch >= '0' && ch <= '9') //得到完整的数
res = ch - '0';
while((ch = getchar()) >= '0' && ch <= '9' )
res = res * 10 + ch - '0';
return flag ? -res : res;
}
/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/
const int MAXN=20010;//点数的最大值
const int MAXM=880010;//边数的最大值
const int INF=0x3f3f3f3f;
struct Node
{
int from,to,next;
int cap;
}edge[MAXM];
int tol;
int head[MAXN];
int dep[MAXN];
int gap[MAXN];//gap[x]=y :说明残留网络中dep[i]==x的个数为y
//n是总的点的个数，包括源点和汇点
void init()
{
tol=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w)
{
edge[tol].from=u;
edge[tol].to=v;
edge[tol].cap=w;
edge[tol].next=head[u];
head[u]=tol++;
edge[tol].from=v;
edge[tol].to=u;
edge[tol].cap=0;
edge[tol].next=head[v];
head[v]=tol++;
}
void BFS(int start,int end)
{
memset(dep,-1,sizeof(dep));
memset(gap,0,sizeof(gap));
gap[0]=1;
int que[MAXN];
int front,rear;
front=rear=0;
dep[end]=0;
que[rear++]=end;
while(front!=rear)
{
int u=que[front++];
if(front==MAXN)front=0;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(dep[v]!=-1)continue;
que[rear++]=v;
if(rear==MAXN)rear=0;
dep[v]=dep[u]+1;
++gap[dep[v]];
}
}
}
int SAP(int start,int end,int n)
{
int res=0;
BFS(start,end);
int cur[MAXN];
int S[MAXN];
int top=0;
memcpy(cur,head,sizeof(head));
int u=start;
int i;
while(dep[start]<n)
{
if(u==end)
{
int temp=INF;
int inser;
for(i=0;i<top;i++)
if(temp>edge[S[i]].cap)
{
temp=edge[S[i]].cap;
inser=i;
}
for(i=0;i<top;i++)
{
edge[S[i]].cap-=temp;
edge[S[i]^1].cap+=temp;
}
res+=temp;
top=inser;
u=edge[S[top]].from;
}
if(u!=end&&gap[dep[u]-1]==0)//出现断层，无增广路
break;
for(i=cur[u];i!=-1;i=edge[i].next)
if(edge[i].cap!=0&&dep[u]==dep[edge[i].to]+1)
break;
if(i!=-1)
{
cur[u]=i;
S[top++]=i;
u=edge[i].to;
}
else
{
int min=n;
for(i=head[u];i!=-1;i=edge[i].next)
{
if(edge[i].cap==0)continue;
if(min>dep[edge[i].to])
{
min=dep[edge[i].to];
cur[u]=i;
}
}
--gap[dep[u]];
dep[u]=min+1;
++gap[dep[u]];
if(u!=start)u=edge[S[--top]].from;
}
}
return res;
}
int emp=0,rownum=0,colnum=0;
struct A
{
int x;
int y;
int val;
}row[15000],col[15000];
int g[200][200];
int print(int tp)
{
int id=tp+rownum;
int ans=0;
for(int i=head[id];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(v<=rownum)
{ans+=edge[i].cap;
break;}
}
return ans+1;
}
int main()
{
int n,m;
while(RII(n,m)!=EOF)
{MS0(g);
init();
char s[10];
emp=0,rownum=0,colnum=0;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
scanf("%s",s);
if(s[0]=='.')
g[i][j]=++emp;
else
{if(s[4]!='X')
{ int v=(s[4]-'0')*100+(s[5]-'0')*10+s[6]-'0';
row[++rownum].x=i;
row[rownum].y=j;
row[rownum].val=v;
}
if(s[0]!='X')
{
int v=(s[0]-'0')*100+(s[1]-'0')*10+s[2]-'0';
col[++colnum].x=i;
col[colnum].y=j;
col[colnum].val=v;
}
}
}
int S=0,T=emp+rownum+colnum+1;
for(int i=1;i<=rownum;i++)
{
int x=row[i].x;
int y=row[i].y;
int v=row[i].val;
int cnt=0;
for(int j=y+1;j<m;j++)
{
if(g[x][j]!=0)
{
cnt++;
addedge(i,rownum+g[x][j],8);
}
else
break;
}
addedge(0,i,v-cnt);
}
for(int i=1;i<=colnum;i++)
{
int x=col[i].x;
int y=col[i].y;
int v=col[i].val;
int cnt=0;
for(int j=x+1;j<n;j++)
{
if(g[j][y])
{
cnt++;
addedge(rownum+g[j][y],rownum+emp+i,8);
}
else
break;
}
addedge(rownum+emp+i,T,v-cnt);
}
SAP(S,T,T+1);
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(g[i][j]==0)
printf("_");
else
{
printf("%d",print(g[i][j]));
}
if(j==m-1)
printf("\n");
else
printf(" ");
}
}
return 0;
}
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