WHU 1124 Football Coach（最大流）

Problem 1124 - Football Coach

Time Limit: 2000MS    Memory Limit: 65536KB    Difficulty: 1 
Total Submit: 82   Accepted: 46   Special Judge: No

Description

It is not an easy job to be a coach of a football team. The season is almost over, only a few matches are left to play. All of sudden the team 
manager comes to you and tells you bad news: the main sponsor of your club is not happy with your results and decided to stop sponsoring your 
team, which probably means the end of your club. The sponsor's decision is final and there is no way to change it unless... unless your team 
miraculously wins the league. 
The manager left you in deep thought. If you increase the number of practices and offer players a generous bonus for each match, you may be 
able to win all the remaining matches. Is that enough? You also have to make sure that teams with many points lose against teams with few 
points so that in the end, your team will have more points than any other team. You know some of the referees and can bribe them to manipulate 
the result of each match. But first you need to figure out how to manipulate the results and whether it is possible at all. 
There are N teams numbered 1 through N, your team has the number N. The current number of points of each team and the list of remaining 
matches are given. Your task is to find out whether it is possible to manipulate each remaining match so that the team N will finish with 
strictly more points than any other team. If it is possible, output "YES", otherwise, output "NO". In every match, the winning team gets 2 
points, the losing team gets 0. If the match ends with a draw, both teams get 1 point. 

Input

There will be multiple test cases. Each test case has the following form: The first line contains two numbers N(1 <= N <= 100) and M(0 <= M <= 
1000). The next line contains N numbers separated by spaces giving the current number of points of teams 1, 2, ..., N respectively. The 
following M lines describe the remaining matches. Each line corresponds to one match and contains two numbers a and b (a not equal to b, 1 <= 
a,b <= N) identifying the teams that will play in the given match. There is a blank line after each test case.

Output

For each test case, output "YES" or "NO" to denote whether it's possible to manipulate the remaining matches so that the team N would win 
the league.

Sample Input

5 8
2 1 0 0 1
1 2
3 4
2 3
4 5
3 1
2 4
1 4
3 5
5 4
4 4 1 0 3
1 3
2 3
3 4
4 5

Sample Output

YES
NO

Hint

The problem is so hard that even I have told you the method here is "maximum network flow", you can't solve it. You can have a try, but don?t waste too much time here if you are not perfect at modeling a network.

Source

2006 Team Select Round 3

题目：http://acm.whu.edu.cn/learn/problem/detail?problem_id=1124

分析：这题题目直接说是最大流了，然后直接想= =

首先贪心，对于跟team N比的比赛，都是team N胜，给他加2分，这样就得到team N 的最好成绩记为limit

对于普通的比赛，开始构图了，把每场比赛和每支队伍看做点，增加源点和汇点，源点与每场比赛连上容量为2的边（每场比赛最多给出2分），每场比赛与他的两个参赛队都连上容量为2的边（这样刚好满足三种结局的分配），然后对于每支队伍与汇点直接，如果这支队伍原有的分数大于limit-1，无解，否则连上容量为（limit-1-原有分数）的边

如果最大流满流就是有解，否则无解

代码：

#include<cstdio>
#include<iostream>
using namespace std;
const int oo=1000000000;
const int mm=6666;
const int mn=1111;
int node,edge,src,dest;
int ver[mm],flow[mm],next[mm];
int head[mn],work[mn],dis[mn],q[mn],score[mn];
void prepare(int _node,int _src,int _dest)
{
    node=_node,src=_src,dest=_dest;
    for(int i=0;i<node;++i)head[i]=-1;
    edge=0;
}

void addedge(int u,int v,int c)
{
    ver[edge]=v,flow[edge]=c,next[edge]=head[u],head[u]=edge++;
    ver[edge]=u,flow[edge]=0,next[edge]=head[v],head[v]=edge++;
}
int Dinic_bfs()
{
    int i,u,v,l,r=0;
    for(i=0;i<node;++i)dis[i]=-1;
    dis[q[r++]=src]=0;
    for(l=0;l<r;++l)
        for(i=head[u=q[l]];i>=0;i=next[i])
            if(flow[i]&&dis[v=ver[i]]<0)
            {
                dis[q[r++]=v]=dis[u]+1;
                if(v==dest)return 1;
            }
    return 0;
}
int Dinic_dfs(int u,int exp)
{
    if(u==dest)return exp;
    for(int &i=work[u],v,tmp;i>=0;i=next[i])
        if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
        {
            flow[i]-=tmp;
            flow[i^1]+=tmp;
            return tmp;
        }
    return 0;
}
int Dinic_flow()
{
    int i,ret=0,delta;
    while(Dinic_bfs())
    {
        for(i=0;i<node;++i)work[i]=head[i];
        if(delta=Dinic_dfs(src,oo))ret+=delta;
    }
    return ret;
}
int main()
{
    int i,u,v,m,n,sum;
    while(scanf("%d%d",&n,&m)!=-1)
    {
        for(i=1;i<=n;++i)scanf("%d",&score[i]);
        prepare(n+m+2,0,n+m+1);
        sum=0;
        for(i=1;i<=m;++i)
        {
            scanf("%d%d",&u,&v);
            if(u==n||v==n)score[n]+=2;
            else
            {
                sum+=2;
                addedge(src,i,2);
                addedge(i,m+u,2);
                addedge(i,m+v,2);
            }
        }
        for(i=1;i<n;++i)
        {
            if(score[i]>=score[n]-1)break;
            addedge(m+i,dest,score[n]-1-score[i]);
        }
        if(i==n&&Dinic_flow()==sum)puts("YES");
        else puts("NO");
    }
    return 0;
}