很裸的判断最小割是否唯一。判断方法是先做一遍最大流求最小割,然后从源点和汇点分别遍历所有能够到达的点,看是否覆盖了所有的点,如果覆盖了所有的点,那就是唯一的,否则就是不唯一的。
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #include <cstdlib> #include <list> #include <set> #include <queue> #include <stack> using namespace std; typedef long long LL; const int maxn = 800 + 5; const int maxm = 80000; const int INF = INT_MAX / 2; int n,m,s,t; int first[maxn],nxt[maxm]; int ecnt,v[maxm],cap[maxm]; int q[maxn],qs,qe; bool vis[maxn]; void adde(int uu,int vv,int ww) { v[ecnt] = vv; cap[ecnt] = ww; nxt[ecnt] = first[uu]; first[uu] = ecnt++; v[ecnt] = uu; cap[ecnt] = 0; nxt[ecnt] = first[vv]; first[vv] = ecnt++; } int level[maxn]; bool bfs() { memset(level,0,sizeof(level)); qs = qe = 0; q[qe++] = s; level[s] = 1; while(qs < qe) { int now = q[qs++]; if(now == t) break; for(int i = first[now];~i;i = nxt[i]) { if(cap[i] && level[v[i]] == 0) { q[qe++] = v[i]; level[v[i]] = level[now] + 1; } } } return level[t]; } int dfs(int now,int alpha) { if(now == t) return alpha; int sum = 0; for(int i = first[now];~i && alpha;i = nxt[i]) { if(cap[i] && level[v[i]] == level[now] + 1) { int ret = dfs(v[i],min(alpha,cap[i])); sum += ret; alpha -= ret; cap[i] -= ret; cap[i ^ 1] += ret; } } if(sum == 0) level[now] = -1; return sum; } void solve() { //先做一遍最大流 while(bfs()) dfs(s,INF); //分别从起点和终点做一遍bfs memset(vis,0,sizeof(vis)); qs = qe = 0; q[qe++] = s; vis[s] = true; while(qs < qe) { int now = q[qs++]; for(int i = first[now];~i;i = nxt[i]) { if(cap[i] && !vis[v[i]]) { vis[v[i]] = true; q[qe++] = v[i]; } } } qs = qe = 0; q[qe++] = t; vis[t] = true; while(qs < qe) { int now = q[qs++]; for(int i = first[now];~i;i = nxt[i]) { if(cap[i ^ 1] && !vis[v[i]]) { vis[v[i]] = true; q[qe++] = v[i]; } } } for(int i = 1;i <= n;i++) { if(!vis[i]) { puts("AMBIGUOUS"); return; } } puts("UNIQUE"); } int main() { while(scanf("%d%d%d%d",&n,&m,&s,&t),n) { memset(first,-1,sizeof(first)); ecnt = 0; for(int i = 0;i < m;i++) { int a,b,c; scanf("%d%d%d",&a,&b,&c); adde(a,b,c); adde(b,a,c); } solve(); } return 0; }
ZOJ 2587 Unique Attack 判断最小割是否唯一,布布扣,bubuko.com
ZOJ 2587 Unique Attack 判断最小割是否唯一
原文:http://www.cnblogs.com/rolight/p/3873313.html