HDU 3572 Task Schedule

Task Schedule

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 3572
64-bit integer IO format: %I64d      Java class name: Main

 

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

Source

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

 

解题：最大流。。建图确实有点。。。一下子想不到。。。还以为要拆边。。。

 

好吧 源点到每个任务建边容量为pi，每个任务到期限内的各天连容量为1的边，每天到汇点连容量为m的边。。。

 

现在很好理解了，如果满流即存在可行方案。。

 

每天最多m台机器同时工作。。。每个任务需要工作pi天。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 20000;
struct arc{
    int to,flow,next;
    arc(int x = 0,int y = 0,int z = -1){
        to = x;
        flow = y;
        next = z;
    }
};
arc e[maxn*10];
int head[maxn],d[maxn],cur[maxn];
int tot,S,T,n,m;
void add(int u,int v,int flow){
    e[tot] = arc(v,flow,head[u]);
    head[u] = tot++;
    e[tot] = arc(u,0,head[v]);
    head[v] = tot++;
}
bool bfs(){
    memset(d,-1,sizeof(d));
    queue<int>q;
    d[T] = 1;
    q.push(T);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        for(int i = head[u]; ~i; i = e[i].next){
            if(e[i^1].flow && d[e[i].to] == -1){
                d[e[i].to] = d[u] + 1;
                q.push(e[i].to);
            }
        }
    }

    return d[S] > -1;
}
int dfs(int u,int low){
    if(u == T) return low;
    int tmp = 0,a;
    for(int &i = cur[u]; ~i; i = e[i].next){
        if(e[i].flow && d[e[i].to] + 1 == d[u]&&(a=dfs(e[i].to,min(e[i].flow,low)))){
            e[i].flow -= a;
            e[i^1].flow += a;
            tmp += a;
            low -= a;
            if(!low) break;
        }
    }
    if(!tmp) d[u] = -1;
    return tmp;
}
int dinic(){
    int ans  = 0;
    while(bfs()){
        memcpy(cur,head,sizeof(head));
        ans += dfs(S,INF);
    }
    return ans;
}
int main() {
    int cs,pi,si,ei,cao = 1;
    scanf("%d",&cs);
    while(cs--){
        scanf("%d %d",&n,&m);
        int mx = 0,sum = 0;
        memset(head,-1,sizeof(head));
        S = tot = 0;
        for(int i = 1; i <= n; ++i){
            scanf("%d %d %d",&pi,&si,&ei);
            sum += pi;
            for(int k = si; k <= ei; ++k) add(i,n+k,1);
            add(S,i,pi);
            mx = max(mx,ei);
        }
        T = n + mx + 1;
        for(int i = n+1; i < T; ++i) add(i,T,m);
        printf("Case %d: %s

",cao++,dinic() == sum?"Yes":"No");
    }
    return 0;
}