hdu 3572 Task Schedule （dinic算法）

pid=3572">Task Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3412    Accepted Submission(s): 1197

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

 

Author

allenlowesy

 

思路：建一个超级源点0，然后如果工作区间长度为T ,再建立[1，T]个点，源点到每一个点的流量为M（每天仅仅有M台机器工作）。接着。把对应的工作日向后平移T 天，每一个工作日到对应的[1,T]的流量为1，到终点的流量也为1.

最后求最大流是否大于等于总总工作量就是了。

#include"stdio.h"
#include"string.h"
#include"queue"
using namespace std;
#define N 1005
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
const int inf=0x7ffffff;
int cnt,n,m,t;
int head[N],q[N],dis[N];
struct node
{
    int u,v,w,next;
}map[N*N];
void add(int u,int v,int w)
{
    map[cnt].u=u;
    map[cnt].v=v;
    map[cnt].w=w;
    map[cnt].next=head[u];
    head[u]=cnt++;
    map[cnt].u=v;
    map[cnt].v=u;
    map[cnt].w=0;
    map[cnt].next=head[v];
    head[v]=cnt++;
}
int bfs()
{
    int i,u,v,t1,t2;
    memset(dis,0,sizeof(dis));
    u=t1=t2=0;
    dis[u]=1;
    q[t1++]=u;
    while(t2<t1)
    {
        u=q[t2++];
        for(i=head[u];i!=-1;i=map[i].next)
        {
            v=map[i].v;
            if(map[i].w&&!dis[v])
            {
                dis[v]=dis[u]+1;
                if(v==t)
                    return 1;
                q[t1++]=v;
            }
        }
    }
    return 0;
}
int dfs(int s,int lim)
{
    int i,tmp,v,cost=0;
    if(s==t)
        return lim;
    for(i=head[s];i!=-1;i=map[i].next)
    {
        v=map[i].v;
        if(map[i].w&&dis[s]==dis[v]-1)
        {
            tmp=dfs(v,min(lim-cost,map[i].w));
            if(tmp>0)
            {
                map[i].w-=tmp;
                map[i^1].w+=tmp;
                cost+=tmp;
                if(cost==lim)
                    break;
            }
            else
                dis[v]=-1;
        }
    }
    return cost;
}
int dinic()
{
    int ans=0,s=0;
    while(bfs())
        ans+=dfs(s,inf);

    //printf("%d\n",ans);
    return ans;
}
int main()
{
    int i,j,T,sum,t1,t2,cas=1;
    int s[505],e[505],p[505];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        t1=N;t2=0;
        sum=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d%d%d",&p[i],&s[i],&e[i]);
            t1=min(t1,s[i]);
            t2=max(t2,e[i]);
            sum+=p[i];
        }
        cnt=0;
        memset(head,-1,sizeof(head));
        for(i=t1;i<=t2;i++)      //超级源点到一般源点的流量
        {
            add(0,i,m);
        }
        for(i=1;i<=n;i++)
        {
            for(j=s[i];j<=e[i];j++)
            {
                add(j,j+t2,1);
                add(j+t2,2*t2,1);
            }
        }
        t=t2*2;
        if(sum<=dinic())
            printf("Case %d: Yes\n\n",cas++);
        else
            printf("Case %d: No\n\n",cas++);
    }
    return 0;
}

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