将一个点拆成3个,分别表示x_scissors , x_rock , x_cloth
如果x>y 那么表示x是scissors y就是cloth , x是rock y就是scissors , x是cloth y就是cloth
合并3个就可以了
我们是通过排除法来确定裁判编号的,枚举每个编号为裁判,一但在某一行输入中出现矛盾,我们就确定它不是裁判,那么n-1个出现矛盾的枚举中出现矛盾最晚的那个行数即为我们能确定裁判所需的最少行数啦(因为我们要排除n-1个编号才能确定裁判的编号嘛)~ ----https://www.cnblogs.com/geloutingyu/p/6145706.html
#include<cstdio>
#define N 505*3
#define M 2005
#define xx find(x)
#define x1 find(x+n)
#define x2 find(x+n+n)
#define yy find(y)
#define y1 find(y+n)
#define y2 find(y+n+n)
using namespace std;
int fa[N],n,m,cnt,Max,pos;
struct Node{int x,y,op;}a[M];
int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
int check(int now){
for(int i=1;i<=m;i++){
int x=a[i].x,y=a[i].y,op=a[i].op;
if(x==now || y==now) continue;
else{
if(op==1){
if(xx!=y1) fa[xx]=y1;
if(x1!=y2) fa[x1]=y2;
if(x2!=yy) fa[x2]=yy;
}
if(op==2){
if(xx!=yy) fa[xx]=yy;
if(x1!=y1) fa[x1]=y1;
if(x2!=y2) fa[x2]=y2;
}
if(op==3){
if(xx!=y2) fa[xx]=y2;
if(x1!=yy) fa[x1]=yy;
if(x2!=y1) fa[x2]=y1;
}
if(xx==x1||xx==x2||x1==x2||yy==y1||y1==y2||yy==y2) return i;
}
}return 0;
}
int main(){
while(~scanf("%d%d",&n,&m)){
cnt=Max=pos=0;
for(int i=1;i<=m;i++){
int x,y; char ch[3];
scanf("%d%c%d",&x,&ch,&y);
a[i].x=x+1 , a[i].y=y+1;
if(ch[0]=='<') a[i].op=1;
if(ch[0]=='=') a[i].op=2;
if(ch[0]=='>') a[i].op=3;
}
for(int i=1;i<=n;i++){
for(int j=1;j<=n*3;j++) fa[j]=j;
int x=check(i);
if(!x) cnt++,pos=i;
else if(x>Max) Max=x;
}
if(cnt==0) printf("Impossible\n");
if(cnt==1) printf("Player %d can be determined to be the judge after %d lines\n",pos-1,Max);
if(cnt>1) printf("Can not determine\n");
}
}