zoj 2587 - Unique Attack（网络流’最小割）

题目：

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1587

题意：

N个点，M条无向边，判断从s到t的最小割是否唯一。

思路：

判断最小割是否唯一的方法是：

跑一次最大流，从源点s和汇点t各一次dfs，用一个数组记录经过的点。

两次dfs之后有点未被记录则最小割不唯一。否则唯一。

证明见：http://www.cnblogs.com/ka200812/archive/2011/07/30/2121872.html

AC.

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <queue>
#include <vector>

using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 805;
int N, M, L;
int vis[maxn];

struct edge {
int to, cap, rev, id;
edge(int t, int c, int r, int d) {
to = t; cap = c; rev = r; id = d;
}
};
vector<edge> g[maxn];
int level[maxn], iter[maxn];

void addedge(int from, int to, int cap, int id)
{
g[from].push_back(edge(to, cap, g[to].size(), id));
g[to].push_back(edge(from, 0, g[from].size()-1, id+1));
}
void bfs(int s)
{
memset(level, -1, sizeof(level));
queue<int> que;
level[s] = 0;
que.push(s);
while(!que.empty()) {
int v = que.front(); que.pop();
for(int i = 0; i < g[v].size(); ++i) {
edge &e = g[v][i];
if(e.cap > 0 && level[e.to] < 0) {
level[e.to] = level[v] + 1;
que.push(e.to);
}
}
}
}
int dfs(int v, int t, int f)
{
if(v == t) return f;
for(int &i = iter[v]; i < g[v].size(); ++i) {
edge &e = g[v][i];
if(e.cap > 0 && level[v] < level[e.to]) {
int d = dfs(e.to, t, min(f, e.cap));
if(d > 0) {
e.cap -= d;
g[e.to][e.rev].cap += d;
return d;
}
}
}
return 0;
}
int max_flow(int s, int t)
{
int flow = 0;
while(1) {
bfs(s);
if(level[t] < 0) return flow;
memset(iter, 0, sizeof(iter));
int f;
while((f = dfs(s, t, INF)) > 0) {
flow += f;
}
}
}
void dfs1(int x)
{
vis[x] = 1;
for(int i = 0; i < g[x].size(); ++i) {
edge e = g[x][i];
if(e.id%2==0 && e.cap && !vis[e.to]) dfs1(e.to);
}
}
void dfs2(int x)
{
vis[x] = 1;
for(int i = 0; i < g[x].size(); ++i) {
edge e = g[x][i];
if(e.id%2==1 && g[e.to][e.rev].cap && !vis[e.to]) dfs2(e.to);
}
}
int main()
{
//freopen("in", "r", stdin);
int s, t;
while(~scanf("%d%d%d%d", &N, &M, &s, &t)) {
if(N == 0 && M == 0 && s == 0 && t == 0) break;

for(int i = 1; i <= maxn; ++i) g[i].clear();
L = M;
for(int i = 0; i < 2*M; i+=2) {
int u, v, c;
scanf("%d%d%d", &u, &v, &c);
addedge(u, v, c, i);
addedge(v, u, c, i+2);
}
int an = max_flow(s, t);
//printf("%d\n", an);

memset(vis, 0, sizeof(vis));
dfs1(s);
dfs2(t);
int ok = 0;
for(int i = 1; i <= N; ++i) {
if(vis[i] == 0) {
ok = 1;
printf("AMBIGUOUS\n");
break;
}
}
if(!ok) printf("UNIQUE\n");
}
return 0;
}