WOJ 124. Football Coach 网络流

Problem 1124 - Football Coach

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

题解：

先将所有和n有关的比赛让n赢

然后再每一场比赛与S相连容量为2

S与有关的队伍相连，容量为2

队伍与T相连容量为Val[n]-Val[i]-1 保证每一个队伍的流量不超过Val[n];

如果流量=2*除去与n有关比赛场数代表所有的比赛都比完了，且没有一个队伍Val>=Val[n] 输出YES

反之如果< 除去与n有关比赛场数代表比赛还没有比完但如果再流就会导致Val>=Val[n] 输出NO

代码：

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<algorithm>
  4 #include<cmath>
  5 #include<cstring>
  6 #include<cstdlib>
  7 using namespace std;
  8 const int N=105,M=10005,INF=1999999999;
  9 int n,m,T;
 10 int val[N];
 11 int gi(){
 12     int str=0;char ch=getchar();
 13     while(ch>'9'||ch<'0')ch=getchar();
 14     while(ch>='0' && ch<='9')str=str*10+ch-'0',ch=getchar();
 15     return str;
 16 }
 17 struct Lin{
 18     int next,to,dis;
 19 }a[M*2];
 20 int head[M],num=1;
 21 void init(int x,int y,int z){
 22     a[++num].next=head[x];
 23     a[num].to=y;
 24     a[num].dis=z;
 25     head[x]=num;
 26     a[++num].next=head[y];
 27     a[num].to=x;
 28     a[num].dis=0;
 29     head[y]=num;
 30 }
 31 int dep[M],q[M],cnt;
 32 bool bfs()
 33 {
 34     memset(dep,0,sizeof(dep));
 35     q[1]=0;dep[0]=1;
 36     int t=0,sum=1,x,u;
 37     while(t!=sum)
 38     {
 39         x=q[++t];
 40         for(int i=head[x];i;i=a[i].next){
 41             u=a[i].to;
 42             if(dep[u] || a[i].dis<=0)continue;
 43             q[++sum]=u;dep[u]=dep[x]+1;
 44         }
 45     }
 46     if(dep[T])return true;
 47     return false;
 48 }
 49 int dfs(int x,int flow)
 50 {
 51     if(x==T || !flow)return flow;
 52     int tot=0,u,tmp;
 53     for(int i=head[x];i;i=a[i].next){
 54         u=a[i].to;
 55         if(dep[u]!=dep[x]+1 || a[i].dis<=0)continue;
 56         tmp=dfs(u,min(flow,a[i].dis));
 57         a[i].dis-=tmp;
 58         a[i^1].dis+=tmp;
 59         flow-=tmp;
 60         tot+=tmp;
 61         if(!flow)break;
 62     }
 63     return tot;
 64 }
 65 bool Flow()
 66 {
 67     int tot=0,a;
 68     while(bfs()){
 69         a=dfs(0,INF);
 70         while(a)tot+=a,a=dfs(0,INF);
 71     }
 72     return tot==cnt*2;
 73 }
 74 void work()
 75 {
 76     T=n+m;cnt=m;
 77     int x,y;
 78     for(int i=1;i<=n;i++)val[i]=gi();
 79     for(int i=1;i<=m;i++){
 80         x=gi();y=gi();
 81         if(x==n || y==n)val[n]+=2,cnt--;
 82         else{
 83             init(0,i,2);
 84             init(i,x+m,2);init(i,y+m,2);
 85         }
 86     }
 87     for(int i=1;i<n;i++){
 88         init(i+m,T,val[n]-val[i]-1);
 89     }
 90     for(int i=1;i<n;i++)if(val[i]>=val[n]){
 91         printf("NO
");
 92         return ;
 93     }
 94     if(Flow())printf("YES
");
 95     else printf("NO
");
 96 }
 97 void Clear()
 98 {
 99     num=1;
100     memset(head,0,sizeof(head));
101 }
102 int main()
103 {
104     while(~scanf("%d%d",&n,&m)){
105         work();
106         Clear();
107     }
108 }