网络流 -- Alice's Chance(二分图多重最大匹配/最大流)

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本文介绍了一个基于二分图多重最大匹配的问题，旨在解决一个演员如何能在限定时间内完成多个电影的拍摄任务。通过构造一个特殊的网络流模型并求解最大流，来确定演员是否能按照要求完成所有电影的拍摄。

解题报告

http://blog.youkuaiyun.com/juncoder/article/details/38237641

题目传送门

题意：

N个电影，每个电影在每一周有固定拍映时间，电影必须在W周前演完。有一个演员，他每天只能演一场电影，对于每部电影必须演完D天才算完。

思路：

二分图多重最大匹配问题，对于每个电影，源点与每个电影连上一条边容量为D，电影与每周7天对应拍映连线，容量为1，每周每天与汇点连线容量为1

在二分图最大匹配中，每个点（不管是X方点还是Y方点）最多只能和一条匹配边相关联，然而，我们经常遇到这种问题，即二分图匹配中一个点可以和多条匹配边相关联，但有上限，或者说，L _i 表示点i最多可以和多少条匹配边相关联。

在原图上建立源点S和汇点T，S向每个X方点连一条容量为该X方点L值的边，每个Y方点向T连一条容量为该Y方点L值的边，原来二分图中各边在新的网络中仍存在，容量为1（若该边可以使用多次则容量大于1），求该网络的最大流，就是该二分图多重最大匹配的值。

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#define inf 99999999
using namespace std;
int mmap[400][400],n,l[400],m;
int bfs()
{
	memset(l,-1,sizeof(l));
	queue<int>Q;
	Q.push(0);
	l[0]=0;
	while(!Q.empty()) {
		int u=Q.front();
		Q.pop();
		for(int i=0; i<=m; i++) {
			if(l[i]==-1&&mmap[u][i]) {
				l[i]=l[u]+1;
				Q.push(i);
			}
		}
	}
	if(l[m]>1)
		return 1;
	return 0;
}
int dfs(int x,int f)
{
	int a;
	if(x==m)return f;
	for(int i=0; i<=m; i++) {
		if(l[i]==l[x]+1&&mmap[x][i]&&(a=dfs(i,min(f,mmap[x][i])))) {
			mmap[x][i]-=a;
			mmap[i][x]+=a;
			return a;
		}
	}
	l[x]=-1;
	return 0;
}
int main()
{
	int t,_h[10],d,w;
	cin>>t;
	while(t--) {
		memset(mmap,0,sizeof(mmap));
		memset(_h,0,sizeof(_h));
		cin>>n;
		int mon=0;
		int sum=0;
		for(int i=1; i<=n; i++) {
			scanf("%d%d%d%d%d%d%d%d%d",&_h[1],&_h[2],&_h[3],&_h[4],&_h[5],&_h[6],&_h[7],&d,&w);
			mmap[0][i]=d;
			sum+=d;
			if(mon<w)
				mon=w;
			for(int j=0; j<w; j++) {
				for(int k=1; k<=7; k++) {
					if(_h[k])
						mmap[i][n+j*7+k]=1;
				}
			}
		}
		m=n+mon*7+1;
		for(int i=n+1; i<=n+mon*7; i++)
			mmap[i][m]=1;
		int ans=0,a;
		while(bfs())
			while(a=dfs(0,inf))
				ans+=a;
		if(sum==ans)
			printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}

Alice's Chance

Time Limit: 1000MS	Memory Limit: 10000K
Total Submissions: 5277	Accepted: 2168

Description

Alice, a charming girl, have been dreaming of being a movie star for long. Her chances will come now, for several filmmaking companies invite her to play the chief role in their new films. Unfortunately, all these companies will start making the films at the same time, and the greedy Alice doesn't want to miss any of them!! You are asked to tell her whether she can act in all the films.

As for a film,

it will be made ONLY on some fixed days in a week, i.e., Alice can only work for the film on these days;
Alice should work for it at least for specified number of days;
the film MUST be finished before a prearranged deadline.

For example, assuming a film can be made only on Monday, Wednesday and Saturday; Alice should work for the film at least for 4 days; and it must be finished within 3 weeks. In this case she can work for the film on Monday of the first week, on Monday and Saturday of the second week, and on Monday of the third week.

Notice that on a single day Alice can work on at most ONE film.

Input

The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case begins with a single line containing an integer N (1 <= N <= 20), the number of films. Each of the following n lines is in the form of "F1 F2 F3 F4 F5 F6 F7 D W". Fi (1 <= i <= 7) is 1 or 0, representing whether the film can be made on the i-th day in a week (a week starts on Sunday): 1 means that the film can be made on this day, while 0 means the opposite. Both D (1 <= D <= 50) and W (1 <= W <= 50) are integers, and Alice should go to the film for D days and the film must be finished in W weeks.

Output

For each test case print a single line, 'Yes' if Alice can attend all the films, otherwise 'No'.

Sample Input

Sample Output

Yes
No

Hint

A proper schedule for the first test case:



date     Sun    Mon    Tue    Wed    Thu    Fri    Sat

week1          film1  film2  film1         film1

week2          film1  film2  film1         film1

week3          film1  film2  film1         film1

week4          film2  film2  film2

Source