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

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

每个机器每台只能执行一个任务，每个任务在同一时段也只能被一台机执行。给每个任务的开始时间和截止时间，和持续天数。最多给500天。

建立超级源点，源点到每个任务的流量为持续时间，每天到超级汇点的流量为M，这样能限制流量，即每天只能只能有M机器工作，然后每个任务到日期内的每一天设置流量为1，限制流量，即每天这个任务最多被一台机器干。欧克收工。

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<queue>

#include<vector>

#define INF 1e9

using namespace std;

const int maxn =1000+10;

struct Edge

{

    int from,to,cap,flow;

    Edge(){}

    Edge(int f,int t,int c,int fl):from(f),to(t),cap(c),flow(fl){}

};

struct Dinic

{

    int n,m,s,t;

    vector<Edge> edges;

    vector<int> G[maxn];

    bool vis[maxn];

    int cur[maxn];

    int d[maxn];

    void init(int n,int s,int t)

    {

        this->n=n, this->s=s, this->t=t;

        edges.clear();

        for(int i=0;i<n;++i) G[i].clear();

    }

    void AddEdge(int from,int to,int cap)

    {

        edges.push_back( Edge(from,to,cap,0) );

        edges.push_back( Edge(to,from,0,0) );

        m=edges.size();

        G[from].push_back(m-2);

        G[to].push_back(m-1);

    }

    bool BFS()

    {

        queue<int> Q;

        memset(vis,0,sizeof(vis));

        vis[s]=true;

        d[s]=0;

        Q.push(s);

        while(!Q.empty())

        {

            int x=Q.front(); Q.pop();

            for(int i=0;i<G[x].size();++i)

            {

                Edge& e=edges[G[x][i]];

                if(!vis[e.to] && e.cap>e.flow)

                {

                    vis[e.to]=true;

                    d[e.to]=d[x]+1;

                    Q.push(e.to);

                }

            }

        }

        return vis[t];

    }

    int DFS(int x,int a)

    {

        if(x==t || a==0) return a;

        int flow=0, f;

        for(int &i=cur[x];i<G[x].size();++i)

        {

            Edge &e=edges[G[x][i]];

            if(d[e.to]==d[x]+1 && (f=DFS(e.to,min(a,e.cap-e.flow) ) )>0)

            {

                e.flow +=f;

                edges[G[x][i]^1].flow -=f;

                flow +=f;

                a -=f;

                if(a==0) break;

            }

        }

        return flow;

    }

    int max_flow()

    {

        int ans=0;

        while(BFS())

        {

            memset(cur,0,sizeof(cur));

            ans +=DFS(s,INF);

        }

        return ans;

    }

}DC;

int full_flow;

int main()

{

    int T; scanf("%d",&T);

    for(int kase=1;kase<=T;++kase)

    {

        int n,m;

        scanf("%d%d",&n,&m);

        full_flow=0;

        int src=0,dst=500+n+1;

        DC.init(500+2+n,src,dst);

        bool vis[maxn];//表示第i天是否被用到

        memset(vis,0,sizeof(vis));

        for(int i=1;i<=n;++i)

        {

            int P,S,E;

            scanf("%d%d%d",&P,&S,&E);

            DC.AddEdge(src,500+i,P);

            full_flow += P;

            for(int j=S;j<=E;++j)

            {

                DC.AddEdge(500+i,j,1);

                vis[j]=true;

            }

        }

        for(int i=1;i<=500;++i)if(vis[i])//被任务覆盖的日子才添加边

            DC.AddEdge(i,dst,m);

        printf("Case %d: %s\n\n",kase,DC.max_flow()==full_flow?"Yes":"No");

    }

    return 0;

}

巴特西

图论--网络流--最大流 HDU 3572 Task Schedule（限流建图，超级源汇）

Problem Description

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