hdu 3572 Task Schedule(邻接表dinic）

Task Schedule

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

 

Source

2010 ACM-ICPC Multi-University Training Contest（13）——Host by UESTC

 

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邻接表dinic没什么好说的，dinic相对于朴素增广路算法就是增加了层次图，每次取消所有同层次点之间的边（为什么是对的呢？因为如果计算完当前层次图的话，如果现在同层次之间的边会成为下次层次图中不同层次的边，然后所有情况都会被计算进去），如果一个点的层次比终点的层次大（怎么办呢？计算完当前层次图之后，如果这个点与终点联通，在不断计算层次图的过程中，迟早会出现只有通过这个点的路还联通终点的情况，这样这个点的层次在这次计算就比终点的层次小了）

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
#define N 1005
const int inf=0x7ffffff;
int cnt,n,m,t;
int head[N],dist[N];
struct node
{
    int u,v,w,next;
}edge[N*N];
int min(int x,int y)
{
    return x>y?y:x;
}
void add(int u,int v,int w)
{
    edge[cnt].u=u;
    edge[cnt].v=v;
    edge[cnt].w=w;
    edge[cnt].next=head[u];
    head[u]=cnt++;
    edge[cnt].u=v;
    edge[cnt].v=u;
    edge[cnt].w=0;
    edge[cnt].next=head[v];
    head[v]=cnt++;
}
int bfs()//寻早新的层次图
{
    int i,u,v;
    queue<int>q;
    memset(dist,0,sizeof(dist));
    u=0;
    dist[u]=1;
    q.push(u);
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        for(i=head[u];i!=-1;i=edge[i].next)
        {
            v=edge[i].v;
            if(edge[i].w&&!dist[v])
            {
                dist[v]=dist[u]+1;
                if(v==t)
                    return 1;
                q.push(v);
            }
        }
    }
    return 0;
}
int dfs(int s,int lim)//lim记录当前路上的最大流
{
    int i,tmp,v,cost=0;
    if(s==t)
        return lim;
    for(i=head[s];i!=-1;i=edge[i].next)
    {
        v=edge[i].v;
        if(edge[i].w&&dist[s]==dist[v]-1)//前往下一个层次
        {
            tmp=dfs(v,min(lim-cost,edge[i].w));
            if(tmp>0)
            {
                edge[i].w-=tmp;
                edge[i^1].w+=tmp;
                cost+=tmp;
                if(cost==lim)
                    break;
            }
            else
                dist[v]=-1;
        }
    }
    return cost;
}
int dinic()
{
    int ans=0,s=0;
    while(bfs())//搜索当前层次图
        ans+=dfs(s,inf);//若层次图存在，则将当前层次图所有可增广的路都增广
    return ans;
}
int main()
{
    freopen("C:\\Users\\Administrator\\Desktop\\input.txt","r",stdin);
    int i,j,T,sum,t1,t2,cas=1;
    int s[N],e[N],p[N];
    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=1;i<=n;i++)
        {
            add(0,i,p[i]);
        }
        t=n+t2-t1+1+1;
        for(i=1;i<=n;i++)
        {
            for(j=s[i];j<=e[i];j++)
            {
                add(i,n+j-t1+1,1);
            }
        }
        for(i=t1;i<=t2;i++)
            add(n+i-t1+1,t,m);
        if(sum==dinic())
            printf("Case %d: Yes\n\n",cas++);
        else
            printf("Case %d: No\n\n",cas++);
    }
    return 0;
}