POJ2912 Rochambeau [扩展域并查集]

　　题目传送门

Rochambeau

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 4463		Accepted: 1545

Description

N children are playing Rochambeau (scissors-rock-cloth) game with you. One of them is the judge. The rest children are divided into three groups (it is possible that some group is empty). You don’t know who is the judge, or how the children are grouped. Then the children start playing Rochambeau game for M rounds. Each round two children are arbitrarily selected to play Rochambeau for one once, and you will be told the outcome while not knowing which gesture the children presented. It is known that the children in the same group would present the same gesture (hence, two children in the same group always get draw when playing) and different groups for different gestures. The judge would present gesture randomly each time, hence no one knows what gesture the judge would present. Can you guess who is the judge after after the game ends? If you can, after how many rounds can you find out the judge at the earliest?

Input

Input contains multiple test cases. Each test case starts with two integers N and M (1 ≤ N ≤ 500, 0 ≤ M ≤ 2,000) in one line, which are the number of children and the number of rounds. Following are M lines, each line contains two integers in [0, N) separated by one symbol. The two integers are the IDs of the two children selected to play Rochambeau for this round. The symbol may be “=”, “>” or “<”, referring to a draw, that first child wins and that second child wins respectively.

Output

There is only one line for each test case. If the judge can be found, print the ID of the judge, and the least number of rounds after which the judge can be uniquely determined. If the judge can not be found, or the outcomes of the M rounds of game are inconsistent, print the corresponding message.

Sample Input

3 3
0<1
1<2
2<0
3 5
0<1
0>1
1<2
1>2
0<2
4 4
0<1
0>1
2<3
2>3
1 0

Sample Output

Can not determine

Player 1 can be determined to be the judge after 4 lines

Impossible

Player 0 can be determined to be the judge after 0 lines

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　　分析：比较复杂的一道扩展域并查集，不仅操作麻烦而且输入输出的要求还贼多。。。做的时候还遇到了一堆玄学错误。。。

　　首先枚举每一个人，表示这个人是裁判，然后将没有这个人参与的比赛情况放入并查集中，如果没有矛盾则说名这个人可以是裁判，否则这个人就不能是裁判。如果发现没有满足要求的人，则输出Impossible，如果裁判不止一个则输出Can not determine，否则就可以输出这个人。在操作的时候可以放一个擂台记录一下line数。当然其中有很多小细节不好一一列举，具体看代码吧。

　　Code：

//It is made by HolseLee on 29th May 2018
//POJ 2912
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<iomanip>
#include<algorithm>
#define Fi(i,a,b) for(int i=a;i<=b;i++)
#define Fx(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int N=;const int M=;
int n,m,rank[N],fa[N];
inline int find(int a)
{
    if(fa[a]!=a){
        int father=find(fa[a]);
        rank[a]=(rank[a]+rank[fa[a]])%;
        fa[a]=father;}
    return fa[a];
}
inline bool check(int a,int b,int c)
{
    int fx=find(a);int fy=find(b);
    if(fx==fy){if((rank[b]-rank[a]+)%!=c)return true;}
    else{fa[fy]=fx;rank[fy]=(rank[a]-rank[b]+c+)%;}return false;
}
int main()
{
    for(;scanf("%d%d",&n,&m)!=EOF;){
        int x[M],y[M],ch[M];
        int tot=,cnt=,ans=,c;bool flag;
        Fi(i,,m){scanf("%d%c%d",&x[i],&ch[i],&y[i]);}
        Fi(i,,n-){flag=true;Fi(j,,n-)fa[j]=j,rank[j]=;
        Fi(j,,m){if(x[j]==i||y[j]==i)continue;
        if(ch[j]=='=')c=;else if(ch[j]=='>')c=;else c=;
        if(check(x[j],y[j],c)){cnt=max(cnt,j);flag=false;break;}}
        if(flag){tot++;ans=i;}}
        if(!tot)printf("Impossible\n");
        else if(tot>)printf("Can not determine\n");
        else printf("Player %d can be determined to be the judge after %d lines\n",ans,cnt);
    }return ;
}