hdu3572 水题最大流 非递归dinic算法

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;

}