ZOJ_2587
这个题目本质上就是去判断最小割是否为1,在做完最大流之后的图上,我们从S出发沿非满流的边能走到的点一定是属于S集合的,其余能够沿非满流边走到T的点一定是属于T集合的,如果这时还没有覆盖所有的点,那么那些点就既可以看作是S集合的,也可以看作是T集合的,自然最小割就不唯一了。
#include<stdio.h> #include<string.h> #include<algorithm> #define MAXD 810 #define MAXM 20010 #define INF 0x3f3f3f3f int N, M, first[MAXD], e, next[MAXM], v[MAXM], flow[MAXM]; int S, T, d[MAXD], q[MAXD], work[MAXD], vis[MAXD]; void add(int x, int y, int z) { v[e] = y, flow[e] = z; next[e] = first[x], first[x] = e ++; } void init() { int i, x, y, z; memset(first, -1, sizeof(first[0]) * (N + 1)), e = 0; for(i = 0; i < M; i ++) { scanf("%d%d%d", &x, &y, &z); add(x, y, z), add(y, x, z); } } int bfs() { int i, j, rear = 0; memset(d, -1, sizeof(d[0]) * (N + 1)); d[S] = 0, q[rear ++] = S; for(i = 0; i < rear; i ++) for(j = first[q[i]]; j != -1; j = next[j]) if(flow[j] && d[v[j]] == -1) { d[v[j]] = d[q[i]] + 1, q[rear ++] = v[j]; if(v[j] == T) return 1; } return 0; } int dfs(int cur, int a) { if(cur == T) return a; for(int &i = work[cur]; i != -1; i = next[i]) if(flow[i] && d[v[i]] == d[cur] + 1) if(int t = dfs(v[i], std::min(a, flow[i]))) { flow[i] -= t, flow[i ^ 1] += t; return t; } return 0; } void DFS(int cur, int k) { int i; vis[cur] = 1; for(i = first[cur]; i != -1; i = next[i]) if(flow[i ^ k] != 0 && !vis[v[i]]) DFS(v[i], k); } void dinic() { while(bfs()) { memcpy(work, first, sizeof(first[0]) * (N + 1)); while(dfs(S, INF)); } } int check() { for(int i = 1; i <= N; i ++) if(!vis[i]) return 0; return 1; } void solve() { dinic(); memset(vis, 0, sizeof(vis[0]) * (N + 1)); DFS(S, 0), DFS(T, 1); printf("%s\n", check() ? "UNIQUE" : "AMBIGUOUS"); } int main() { while(scanf("%d%d%d%d", &N, &M, &S, &T), N) { init(); solve(); } return 0; }