Rochambeau
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 5208 | Accepted: 1778 |
题目链接：http://poj.org/problem?id=2912
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
题意：
n个小朋友被分为三组玩剪刀石头布，每一组的小朋友只出固定的招数，但是这些小朋友里面有个裁判，他可以任意出，现在问是否能够确定出这个裁判，最少经过几轮？
题解：
三个组并且是个关于输赢的环状关系，所以我们可以思考带权并查集，v[x]代表x与其父亲结点的关系，v[x]=0代表同类，v[x]=1代表x赢他父亲，v[x]=2就代表输了。
现在关键是这个裁判，我们通过思考可以发现，假如确定了这个人为裁判，那么他参与的几次都不算做分组。
但如何去确定就是个问题。注意这题时间限制挺大的，所以我们直接枚举，假定每个人都作为裁判就行了。
最后还要确定经过几轮游戏，这里挺有意思的：
确定目前这个人为裁判，这等价于排除其他所有人不是裁判的最少次数，这个在枚举其他人作为裁判时就可以进行处理了。
代码如下：
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> using namespace std; const int N = 2505; int n,m; int f[N],a[N],b[N],v[N],ans[N]; char c[N]; int find(int x){ if(x==f[x]) return x; int tmp=f[x]; f[x]=find(f[x]); v[x]=(v[x]+v[tmp])%3; return f[x]; } int main(){ while(~scanf("%d%d",&n,&m)){ for(int i=1;i<=m;i++) scanf("%d%c%d",&a[i],&c[i],&b[i]); int tot = 0,cnt = 0,l = 0; memset(ans,0,sizeof(ans)); for(int i=0;i<n;i++){ bool flag=true;cnt=0; for(int j=0;j<=n;j++) f[j]=j,v[j]=0; for(int j=1;j<=m;j++){ if(a[j]==i || b[j]==i) continue; int fx=find(a[j]),fy=find(b[j]); if(c[j]=='>'){ if(fx==fy && (v[a[j]]+3-v[b[j]])%3!=2) flag=false,cnt=j; else{ f[fx]=fy; v[fx]=(3-v[a[j]]+2+v[b[j]])%3; } }else if(c[j]=='<'){ if(fx==fy && (v[a[j]]+3-v[b[j]])%3!=1) flag=false,cnt=j; else{ f[fx]=fy; v[fx]=(3-v[a[j]]+1+v[b[j]])%3; } }else{ if(fx==fy && (v[a[j]]+3-v[b[j]])%3!=0) flag=false,cnt=j; else{ f[fx]=fy; v[fx]=(3-v[a[j]]+v[b[j]])%3; } } if(!flag) break; } if(flag) ans[++tot]=i; l=max(l,cnt); } if(tot==0) puts("Impossible"); else if(tot>1) puts("Can not determine"); else printf("Player %d can be determined to be the judge after %d lines ",ans[1],l); } return 0; }