ZOJ@2587题意：判断最小割是否唯一。
思路（转）：判断最小割是否唯一，先求一次最大流，然后在残留网络中分别从源汇开始dfs一次，找出最小割[S,T]，如果SUT不包含所有点，那么最小割不唯一。假设点i不被SUT包含，那么残留网络中s不能到达i,i不能到达t,即进入i的边和从i出去的边都满流，假设某条进入i的边x满流，这些流量从若干条边y流出i，那么，如果选x为割边，或者选所有对应的y为割边，不会影响最大流，即最小割容量不变，最小割也就不唯一。
// 2390377 2011-01-21 16:52:22 Accepted 2587 C++ 150 3324 redsea
// Dinic最大流 O(V^2 * E)
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define N 802
#define E 200050
#define typec int // type of cost
const typec inf = 30000000; // max of cost
struct edge {
int x, y, nxt; typec c;
}bf[E];
int ne, head[N], cur[N], ps[N], dep[N];
void addedge(int x, int y, typec c)
{
// add an arc(x -> y, c); vertex: 0 ~ n-1;
bf[ne].x = x; bf[ne].y = y; bf[ne].c = c;
bf[ne].nxt = head[x]; head[x] = ne++;
bf[ne].x = y; bf[ne].y = x; bf[ne].c = 0;
bf[ne].nxt = head[y]; head[y] = ne++;
}
typec flow(int n, int s, int t)
{
typec tr, res = 0;
int i, j, k, f, r, top;
while (1) {
memset(dep, -1, n * sizeof(int));
for (f = dep[ps[0] = s] = 0, r = 1; f != r; )
for (i = ps[f++], j = head[i]; j; j = bf[j].nxt)
{
if (bf[j].c && -1 == dep[k = bf[j].y]){
dep[k] = dep[i] + 1; ps[r++] = k;
if (k == t) { f = r; break; }
}
}
if (-1 == dep[t]) break;
memcpy(cur, head, n * sizeof(int));
for (i = s, top = 0; ; ) {
if (i == t) {
for (k = 0, tr = inf; k < top; ++k)
if (bf[ps[k]].c < tr)
tr = bf[ps[f = k]].c;
for (k = 0; k < top; ++k)
bf[ps[k]].c -= tr, bf[ps[k]^1].c += tr;
res += tr; i = bf[ps[top = f]].x;
}
for (j=cur[i]; cur[i]; j = cur[i] = bf[cur[i]].nxt)
if (bf[j].c && dep[i]+1 == dep[bf[j].y])break;
if (cur[i]) {
ps[top++] = cur[i];
i = bf[cur[i]].y;
}
else {
if (0 == top) break;
dep[i] = -1; i = bf[ps[--top]].x;
}
}
}
return res;
}
int cnts, cntt;
int flag1[N], flag2[N];
void dfs1(int v)
{
flag1[v] = 1;
cnts++;
for(int i = head[v]; i != 0; i = bf[i].nxt){
if(flag1[bf[i].y]==0 && bf[i].c)
dfs1(bf[i].y);
}
}
void dfs2(int v)
{
flag2[v] = 1;
cntt++;
for(int i = head[v]; i != 0; i = bf[i].nxt){
if(flag2[bf[i].y]==0 && bf[i^1].c)
dfs2(bf[i].y);
}
}
int main()
{
int n, m, s, t, x, y, c;
while(scanf("%d%d%d%d",&n,&m,&s,&t), n||m||s||t)
{
memset(head,0,sizeof(head));
ne = 2;
for(int i = 0; i < m; i++){
scanf("%d%d%d",&x,&y,&c);
x--,y--;
addedge(x,y,c);
addedge(y,x,c);
}
s--;t--;
flow(n,s,t);
memset(flag1,0,sizeof(flag1));
memset(flag2,0,sizeof(flag2));
cnts = cntt = 0;
dfs1(s);
dfs2(t);
if(cnts+cntt!=n)
printf("AMBIGUOUS\n");
else printf("UNIQUE\n");
}
return 0;
}