Task Schedule HDU

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.
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.
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.
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
用ek算法无限t，学了dinic好多了

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<queue>#define INF 0x3f3f3f3fusing namespace std;const int maxn=1505;const int maxx=520010;int edge;int to[maxx],flow[maxx],nex[maxx];int head[maxn];void addEdge(int v,int u,int cap){    to[edge]=u,flow[edge]=cap,nex[edge]=head[v],head[v]=edge++;    to[edge]=v,flow[edge]=0,nex[edge]=head[u],head[u]=edge++;}int vis[maxn];int pre[maxn];bool bfs(int s,int e){    queue<int> que;    pre[s]=-1;    memset(vis,-1,sizeof(vis));    que.push(s);    vis[s]=0;    while(!que.empty())    {        int u=que.front();        que.pop();        for(int i=head[u];~i;i=nex[i])        {            int v=to[i];            if(vis[v]==-1&&flow[i])            {                vis[v]=vis[u]+1;                if(v==e)                    return true;                que.push(v);            }        }    }    return false;}int dfs(int s,int t,int f){    if(s==t||!f)        return f;    int r=0;    for(int i=head[s];~i;i=nex[i])    {        int v=to[i];        if(vis[v]==vis[s]+1&&flow[i])        {            int d=dfs(v,t,min(f,flow[i]));            if(d>0)            {                flow[i]-=d;                flow[i^1]+=d;                r+=d;                f-=d;                if(!f)                    break;            }        }    }    if(!r)        vis[s]=INF;    return r;}int maxFlow(int s ,int e){    int ans=0;    while(bfs(s,e))        ans+=dfs(s,e,INF);    return ans;}void init(){    memset(head,-1,sizeof(head));    edge=0;}int main(){    int t;    scanf("%d",&t);    int n,m;    int p,s,e;    int top=-1;    int cal=1;    while(t--)    {        scanf("%d%d",&n,&m);        int sum=0;        init();        for(int i=1;i<=n;i++)        {            scanf("%d%d%d",&p,&s,&e);            sum+=p;            top=max(top,e);            addEdge(0,i,p);            for(int j=s;j<=e;j++)                addEdge(i,n+j,1);        }        for(int i=1;i<=top;i++)            addEdge(n+i,top+n+1,m);        if(sum==maxFlow(0,top+n+1))            printf("Case %d: Yes\n\n",cal++);        else            printf("Case %d: No\n\n",cal++);    }    return 0;}