首页 > 代码库 > HDU 3572 最大流

HDU 3572 最大流

2024-09-28 23:05:01   214人阅读

Task Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8610    Accepted Submission(s): 2636


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
 题意：
有n个任务，m台机器，给出每个任务需要耗费的时间以及能做此任务的时间点，每个任务每天只能由一台机器做，一台机器每天只能做一个任务问能否将这些任务做完
输入n，m
输入n行x，y，z表示该任务需要x天，第y天到第z天可以做这个任务
代码：
//从原点连向n个任务容量为该任务需要的时间，n个任务连向500个时间点容量为1(一个任务一天只有一台机器去做)，//时间点连向汇点，容量为机器数m（一台机器一天只能做一个任务），求最大流是否等于完成所有任务的天数。#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<queue>using namespace std;const int maxn=2009;const int inf=0x7fffffff;struct Edge{    int from,to,cap,flow;    Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}};struct Dinic{    int n,m,s,t;    vector<Edge>edges;    vector<int>g[maxn];    bool vis[maxn];    int d[maxn];    int cur[maxn];    void Init(int n){        this->n=n;        for(int i=0;i<n;i++) g[i].clear();        edges.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));//反向弧        m=edges.size();        g[from].push_back(m-2);        g[to].push_back(m-1);    }    bool Bfs(){        memset(vis,0,sizeof(vis));        queue<int>q;        q.push(s);        d[s]=0;        vis[s]=1;        while(!q.empty()){            int x=q.front();q.pop();            for(int i=0;i<(int)g[x].size();i++){                Edge &e=edges[g[x][i]];                if(!vis[e.to]&&e.cap>e.flow){                    vis[e.to]=1;                    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<(int)g[x].size();i++){            Edge &e=edges[g[x][i]];            if(d[x]+1==d[e.to]&&(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 Maxflow(int s,int t){        this->s=s;this->t=t;        int flow=0;        while(Bfs()){            memset(cur,0,sizeof(cur));            flow+=Dfs(s,inf);        }        return flow;    }}dc;int main(){    int t,n,m;    scanf("%d",&t);    for(int cas=1;cas<=t;cas++){        scanf("%d%d",&n,&m);        dc.Init(n+502);        int x,y,z,sum=0;        for(int i=1;i<=n;i++){            scanf("%d%d%d",&x,&y,&z);            sum+=x;            dc.Addedge(0,i,x);            for(int j=y;j<=z;j++)                dc.Addedge(i,j+n,1);        }        for(int i=1;i<=500;i++)            dc.Addedge(i+n,501+n,m);        printf("Case %d: ",cas);        int tmp=dc.Maxflow(0,501+n);        if(sum==tmp) printf("Yes\n\n");        else printf("No\n\n");    }    return 0;}

 

 

HDU 3572 最大流


联系
我们