【WOJ 124】Football Coach

【题目】

传送门

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. Thefollowing 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.

【分析】

大致题意：给出 $n$ 个球队，每个球队有一个初始分数，再给出 $m$ 场比赛，每次比赛给出两个参赛的球队。每场比赛，赢的队伍得两分，平局两队各得一分，输的得零分。现在你可以决定每场比赛的胜负，问是否有一种方案，使得最后第 $n$ 个球队的分数最高。

最大流板题

如果一个比赛中出现了 $n$，那我们肯定贪心让 $n$ 赢，让 $n$ 得 2 分；不然的话我们相当于是在两个队伍中分配这 2 分，那就变成了一道最大匹配的题

还是细讲一下连边：

1. 连边的顺序是"源点—比赛—球队—汇点"
2. 从源点向每场比赛连一条容量为 2 的边，表示这场比赛总共有 2 分可以分配
3. 从每场比赛向参赛的球队分别连一条容量为 2 的边，表示分配这 2 分
4. 从每个球队向汇点连一条 $scor{e}_n-scor{e}_i-1$ 的边，表示它最多分配这么多分（不然就超过 $n$ 的分数）
5. 最后看跑出来最大流是否能匹配完所有比赛即可

【代码】

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 1005
#define M 10005
#define inf (1ll<<31ll)-1
using namespace std;
int n,m,s,t,num;
int v[M],w[M],nxt[M];
int d[N],f[N],first[N],score[N];
void add(int x,int y,int z)
{
	num++;
	nxt[num]=first[x];
	first[x]=num;
	v[num]=y;
	w[num]=z;
}
bool bfs()
{
	int x,y,i,j;
	memset(d,-1,sizeof(d));
	memcpy(f,first,sizeof(f));
	queue<int>q;d[s]=0;q.push(s);
	while(!q.empty())
	{
		x=q.front();
		q.pop();
		for(i=first[x];i;i=nxt[i])
		{
			y=v[i];
			if(w[i]&&d[y]==-1)
			{
				d[y]=d[x]+1;
				if(y==t)
				  return true;
				q.push(y);
			}
		}
	}
	return false;
}
int dinic(int now,int flow)
{
	if(now==t)  return flow;
	int x,delta,ans=0;
	for(int &i=f[now];i;i=nxt[i])
	{
		x=v[i];
		if(w[i]&&d[x]==d[now]+1)
		{
			delta=dinic(x,min(flow,w[i]));
			w[i]-=delta,w[i^1]+=delta;
			flow-=delta,ans+=delta;
			if(!flow)  return ans;
		}
	}
	return ans;
}
int main()
{
	int x,y,i,j,ans=0;
	while(~scanf("%d%d",&n,&m))
	{
		num=1;
		memset(first,0,sizeof(first));
		int match=0;
		ans=0,s=0,t=m+n+1;
		for(i=1;i<=n;++i)
		  scanf("%d",&score[i]);
		for(i=1;i<=m;++i)
		{
			scanf("%d%d",&x,&y);
			if(x==n||y==n)
			  score[n]+=2;
			else
			{
				match++;
				add(s,i,2),add(i,s,0);
				add(i,m+x,2),add(m+x,i,0);
				add(i,m+y,2),add(m+y,i,0);
			}
		}
		for(i=1;i<n;++i)
		{
			if(score[i]>score[n])
			{
				printf("NO\n");
				break;
			}
			add(m+i,t,score[n]-score[i]-1);
			add(t,m+i,0);
		}
		if(i<n)  continue;
		while(bfs())
		  ans+=dinic(s,inf);
		if(ans/2<match)  printf("NO\n");
		else  printf("YES\n");
	}
	return 0;
}