Task Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4624    Accepted Submission(s): 1516

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.

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

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
#include<climits>
#define Max(a,b) a>b?a:b
#define Min(a,b) a<b?a:b
using namespace std;
struct Edge
{
    int s,t,f,next;
}edge[1100000];
int head[1010];
int cur[1010];
int pre[1010];
int stack[1100000];
int ent;
int n,m,times,s,t;
void add(int start,int last,int f)
{
    edge[ent].s=start;edge[ent].t=last;edge[ent].f=f;edge[ent].next=head[start];head[start]=ent++;
    edge[ent].s=last;edge[ent].t=start;edge[ent].f=0;edge[ent].next=head[last];head[last]=ent++;
}
bool bfs(int S,int T)
{
    memset(pre,-1,sizeof(pre));
    pre[S]=0;
    queue<int>q;
    q.push(S);
    while(!q.empty())
    {
        int temp=q.front();
        q.pop();
        for(int i=head[temp];i!=-1;i=edge[i].next)
        {
            int temp2=edge[i].t;
            if(pre[temp2]==-1&&edge[i].f)
            {
                pre[temp2]=pre[temp]+1;
                q.push(temp2);
            }
        }
    }
    return pre[T]!=-1;
}
int dinic(int start,int last)
{
    int flow=0,now;
    while(bfs(start,last))
    {
        int top=0;
        memcpy(cur,head,sizeof(head));
        int u=start;
        while(1)
        {
            if(u==last)//如果找到终点结束对中间路径进行处理并计算出该流
            {
                int minn=INT_MAX;
                for(int i=0;i<top;i++)
                {
                    if(minn>edge[stack[i]].f)
                    {
                        minn=edge[stack[i]].f;
                        now=i;
                    }
                }
                flow+=minn;
                for(int i=0;i<top;i++)
                {
                    edge[stack[i]].f-=minn;
                    edge[stack[i]^1].f+=minn;
                }
                top=now;
                u=edge[stack[top]].s;
            }
            for(int i=cur[u];i!=-1;cur[u]=i=edge[i].next)//找出从u点出发能到的边
                if(edge[i].f&&pre[edge[i].t]==pre[u]+1)
                    break;
            if(cur[u]==-1)//如果从该点未找到可行边，将该点标记并回溯
            {
                if(top==0)break;
                pre[u]=-1;
                u=edge[stack[--top]].s;
            }
            else//如果找到了继续运行
            {
                stack[top++]=cur[u];
                u=edge[cur[u]].t;
            }
        }
    }
    return flow;
}
int main()
{
    scanf("%d",&times);
    for(int cas=1;cas<=times;cas++)
    {
        ent=0;
        memset(head,-1,sizeof(head));
        int st,ed,lt;
        int maxn=0;
        int sum=0;
        s=0;t=1001;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d%d",&lt,&st,&ed);
            maxn=Max(maxn,ed);
            for(int j=st;j<=ed;j++)
                add(j,i+500,1);
            add(i+500,t,lt);
            sum+=lt;
        }
        for(int i=1;i<=maxn;i++)
            add(s,i,m);
        if(dinic(s,t)==sum)printf("Case %d: Yes\n",cas);
        else printf("Case %d: No\n",cas);
        printf("\n");
    }
    return 0;
}