题目大意:给出N个人剪刀石头布的结果,问能否判断哪个人是裁判,在第几个回合可以判定
解题思路:如果这个人是裁判的话,那么去掉它的话,对结果是不影响的
所以可以暴力枚举一下谁当裁判,然后判断一下
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 510
#define maxm 2010
int f[maxn], r[maxn], p1[maxm], p2[maxm], sign[maxm], n, m;
void input() {
char c;
for(int i = 0; i < m; i++) {
scanf("%d",&p1[i]);
while((c = getchar()) == ' ');
scanf("%d",&p2[i]);
if(c == '<')
sign[i] = 1;
else if(c == '>')
sign[i] = 2;
else
sign[i] = 0;
}
}
void init() {
for(int i = 0; i <= n; i++) {
f[i] = i;
r[i] = 0;
}
}
int find(int x) {
if(x == f[x])
return x;
int t = find(f[x]);
r[x] = (r[x] + r[f[x]]) % 3;
return f[x] = t;
}
void solve() {
int ans = -1, line = 0;
for(int i = 0; i < n; i++) {
init();
int tmp = -1, j;
for(j = 0; j < m; j++) {
int x = p1[j], y = p2[j];
if(x == i || y == i)
continue;
int xx = find(x);
int yy = find(y);
if(xx == yy) {
if( ((3 - r[y] + r[x]) % 3) != sign[j]) {
tmp = j + 1;
break;
}
}
else {
f[xx] = yy;
r[xx] = (r[y] + sign[j] - r[x] + 3) % 3;
}
}
if(j == m) {
if(ans != -1) {
printf("Can not determine\n");
return ;
}
else
ans = i;
}
else
line = max(line,tmp);
}
if(ans == -1)
printf("Impossible\n");
else
printf("Player %d can be determined to be the judge after %d lines\n", ans, line);
}
int main() {
while(scanf("%d%d", &n, &m) == 2){
input();
solve();
}
return 0;
}