(最大流) POJ 1698 Alice's Chance

Alice's Chance

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1613		Accepted: 740

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,

1. it will be made ONLY on some fixed days in a week, i.e., Alice can only work for the film on these days;
2. Alice should work for it at least for specified number of days;
3. 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

2
2
0 1 0 1 0 1 0 9 3
0 1 1 1 0 0 0 6 4
2
0 1 0 1 0 1 0 9 4
0 1 1 1 0 0 0 6 2

Sample Output

YesNo

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

/*
题目大意：有人找Alice拍电影,Alice为了成为明星相抓住所有机会,在导演的要求下拍摄完所有的电影.问可不可以？
		：每个电影必须在他的拍摄周期内的特定日子内才可以拍摄.
题目解答：构图题。只要构造出图，然后用最大流求解就可以解决.建立原点s节点0，film节点1-max_film，
		：然后day节点max_film-max_film*7，汇点t节点max_week-max_film*7+1.
		：s->film[i]==film[i].days，film[j]->day[i]==1,day[i]->t=1;到此图构造完成.
代码技巧：用bfs求一条增益路径,用next[i]存储那几个day是可以利用的.没必要重复判断.

Source Code

Problem: 1698  User: wawadimu 
Memory: 1316K  Time: 782MS 
Language: C++  Result: Accepted 

Source Code
*/ 
#include<iostream>
#include<queue>
using namespace std;

#define maxn 22
#define maxl 380
#define inf INT_MAX
struct node
{
	int next[7];//next[i]下一个可以工作的日子
	int d,w;//天数和星期
}f[maxn];

int cap[maxl][maxl],flow[maxl][maxl];
int a[maxl],pre[maxl];

int maxflow(int s,int t)
{
	int max=0;
	while(1)
	{
		memset(a,0,sizeof(a));
		a[s]=inf;
		queue<int> q;
		q.push(s);
		while(!q.empty())//寻找增益路径
		{
			int u=q.front(); q.pop();
			for(int v=0;v<=t;v++)
			{
				if(!a[v] && cap[u][v] > flow[u][v])
				{
					pre[v]=u;
					a[v]=a[u] > cap[u][v]- flow[u][v] ? cap[u][v]- flow[u][v] : a[u];
					q.push(v);
				}
			}
		}
		if(a[t]==0) break;//没有增益路径
		for(int i=t;i!=s;i=pre[i])//反向增加流量
		{
			flow[pre[i]][i]+=a[t];
			flow[i][pre[i]]-=a[t];
		}
		max+=a[t];
	}
	return max;
}
int main()
{
	//freopen("1698.txt","r",stdin);
	int Case,fn;
	int i,j,k,max,sum;
	int s,t;
	int p,tmp;
	scanf("%d",&Case);
	while(Case--)
	{
		memset(cap,0,sizeof(cap));
		memset(flow,0,sizeof(flow));
		max=sum=0;
		scanf("%d",&fn);
		for(i=1;i<=fn;i++)
		{
			k=0;
			for(j=1;j<=7;j++)
			{
				scanf("%d",&tmp);
				if(tmp==1) f[i].next[k++]=j;//记录可以工作的日期
			}
			scanf("%d%d", &f[i].d, &f[i].w);
			sum+=f[i].d;
			cap[0][i]=f[i].d;
			if(max < f[i].w) max=f[i].w;//记录最大星期
			tmp=f[i].w;
			for(p=0;p<tmp;p++)//将前w个星期每一个可以工作的日子容量设为1
			{
				int kk=7*p+fn;
				for(j=0;j<k;j++)
				{
					int xx=kk+f[i].next[j];
					cap[i][xx]=1;
				}
			}	
		}
		s=0;
		t=7*max+fn+1;
		for(j=fn+1;j<t;j++)//日期到汇点的容量设为1
		{
			cap[j][t]=1;
		}
		int ans=maxflow(0,t);//求最大流
		//cout<<ans<<endl;;
		if(ans==sum)//可以完成所有电影的拍摄
			printf("Yes");
		else
			printf("No");
		printf("/n");
	}
	return 1;
}