POJ 2912 Rochambeau

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并查集应用。

这道题目和 POJ 1182 食物链非常类似。

枚举所有judge，不处理出现当前judge的round，然后利用食物链的解法判断冲突。就可以找出来最终的judge。

还有一步就是找到judge之后，要得到确定此judge的round数：这个round数就是在枚举的时候，其他人出错的round数的最大值——因为，到这个最大round数时，其他人都被排除，剩下的就是真正的judge

#include <cstdio>#include <cstring>const int MAXN = 505;const int MAXM = 2004;int set[MAXN], a[MAXN];struct node {    int u, v;    char ch;} p[MAXM];/*    a[x] = 0 =    a[x] = 1 <    a[x] = 2 >*/int f1[3][3] = {{0, 1, 2}, {2, 0, 1}, {1, 2, 0}};int f2[3][3] = {{1, 2, 0}, {0, 1, 2}, {2, 0, 1}};int f3[3][3] = {{2, 0, 1}, {1, 2, 0}, {0, 1, 2}};int n, m;int Find_set(int x) {    if (set[x] == x) return x;    int t = Find_set(set[x]);    a[x] = (a[x] + a[set[x]]) % 3;    return set[x] = t;}void Union(int x, int y, char ch) {    int fx = Find_set(x);    int fy = Find_set(y);    set[fx] = fy;    if (ch == '=') a[fx] = f1[a[x]][a[y]];    else if (ch == '<') a[fx] = f2[a[x]][a[y]];    else a[fx] = f3[a[x]][a[y]];}void solve() {    int x, y, fx, fy;    int err[MAXN];    memset(err, 0, sizeof(err));    for (int j=0; j<n; j++) {  //枚举judge        for (int i=0; i<n; i++)            set[i] = i, a[i] = 0;        for (int i=0; i<m; i++) {            x = p[i].u, y = p[i].v;            if (x == j || y == j) continue;            fx = Find_set(x);            fy = Find_set(y);            if (fx == fy) {                if (p[i].ch == '=' && a[x]!=a[y]) {                    err[j] = i+1; break;                }                if (p[i].ch == '<' && (a[x]+2)%3 != a[y]) {                    err[j] = i+1; break;                }                if (p[i].ch == '>' && (a[x]+1)%3 != a[y]) {                    err[j] = i+1; break;                }            } else Union(x, y, p[i].ch);        }    }    int cnt = 0, k, ans = 0;    for (int i=0; i<n; i++) {        if (err[i] == 0)  cnt++, k = i;        if (err[i] > ans) ans = err[i];    }    if (cnt == 0) printf("Impossible\n");    else if (cnt == 1)        printf("Player %d can be determined to be the judge after %d lines\n", k, ans);    else printf("Can not determine\n");}int main() {        while (scanf("%d %d", &n, &m) == 2) {        for (int i=0; i<m; i++)            scanf(" %d%c%d", &p[i].u, &p[i].ch, &p[i].v);        solve();    }    return 0;}