2018.07.06 POJ1698 Alice's Chance（最大流）

Alice’s Chance

Time Limit: 1000MS Memory Limit: 10000K

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

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

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

POJ Monthly–2004.07.18

一眼题，显然是简单最大流。这题要我们做的就是限制Alice" role="presentation" style="position: relative;">AliceAlice每部电影的制作次数和每一天的使用次数，以及要保证Alice" role="presentation" style="position: relative;">AliceAlice可以在不同的周选择同样的日子，那么我们将每一周的每一天当做一个点，分别让每个任务和它可以被制作的日子建边（分别与第一周的星期i" role="presentation" style="position: relative;">ii，第二周的星期i" role="presentation" style="position: relative;">ii……第w−1" role="presentation" style="position: relative;">w−1w−1周的星期i" role="presentation" style="position: relative;">ii连容量为1" role="presentation" style="position: relative;">11的边来保证每一天都只能被算一次），对于每一个任务，我们将源点与它们分别连边，容量为di" role="presentation" style="position: relative;">didi，对于每一个日子，我们将它们与汇点t" role="presentation" style="position: relative;">tt分别连边，容量为1" role="presentation" style="position: relative;">11，然后对整张图跑最大流，最后检查是不是每条从源点出发的边的剩余容量都为0" role="presentation" style="position: relative;">00就行了。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#define N 2000
#define M 100005
using namespace std;
int first[N],n,m,s,t,cnt,d[N],T,f[8];
struct edge{int v,next,c;}e[M];
inline void add(int u,int v,int c){
    e[++cnt].v=v;
    e[cnt].c=c;
    e[cnt].next=first[u];
    first[u]=cnt;
    e[++cnt].v=u;
    e[cnt].c=0;
    e[cnt].next=first[v];
    first[v]=cnt;
}
inline bool bfs(){
    queue<int>q;
    memset(d,-1,sizeof(d));
    q.push(s),d[s]=0;
    while(!q.empty()){
        int x=q.front();
        q.pop();
        for(int i=first[x];i!=-1;i=e[i].next){
            int v=e[i].v;
            if(d[v]!=-1||e[i].c<=0)continue;
            d[v]=d[x]+1;
            if(v==t)return true;
            q.push(v);
        }
    }
    return false;
}
inline int dfs(int x,int f){
    if(x==t||!f)return f;
    int flow=f;
    for(int i=first[x];i!=-1;i=e[i].next){
        int v=e[i].v;
        if(d[v]==d[x]+1&&e[i].c>0&&flow){
            int tmp=dfs(v,min(flow,e[i].c));
            if(!tmp)d[v]=-1;
            e[i].c-=tmp;
            e[i^1].c+=tmp;
            flow-=tmp;
        }
    }
    return f-flow;
}
inline int read(){
    int ans=0;
    char ch=getchar();
    while(!isdigit(ch))ch=getchar();
    while(isdigit(ch))ans=(ans<<3)+(ans<<1)+ch-'0',ch=getchar();
    return ans;
}
int main(){
    T=read();
    while(T--){
        n=read(),s=0,cnt=-1;
        int maxn=0,ans=0;
        memset(first,-1,sizeof(first));
        memset(e,0,sizeof(e));
        for(int i=1;i<=n;++i){
            for(int j=1;j<=7;++j)f[j]=read();
            int d=read(),w=read();
            maxn=max(maxn,w);
            add(s,i,d);
            for(int j=1;j<=7;++j)if(f[j])for(int k=0;k<w;++k)add(i,n+j+k*7,1);
        }
        t=n+1+7*maxn;
        for(int i=1;i<=7;++i)
            for(int j=0;j<maxn;++j)
                add(n+i+j*7,t,1);
        while(bfs())ans+=dfs(s,0x3f3f3f3f);
        bool f=true;
        for(int i=first[s];i!=-1;i=e[i].next)if(e[i].c>0){f=false;break;}
        if(f)puts("Yes");
        else puts("No");
    }
    return 0;
}