题意
判断一个无向图的割是否唯一
思路
错误思路：一开始想的是判断割边是否都是关键割边，那既然割边两端点能连通S、T点的边是关键边，那么只要遇到有某个边两端点不连通S or T则这条边就不是关键割边（当然要把不是割边的满流边筛掉）。这种主观臆断的naive想法是不行的，因为那个判断关键割边的条件只是个充分条件，我们没法证明它是个充要条件。
正确思路：求完最大流后在残留网络中从S点开始dfs遍历，再把残留网络反向，然后从T点开始dfs遍历，如果能遍历到所有点则最小割唯一。
代码
#include
#include
#include
#include
#include
#include
#define MID(x,y) ((x+y)/2) #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; const int MAXV = 1005; const int MAXE = 30005; const int oo = 0x3fffffff; struct node{ int u, v, flow; int opp; int next; }; struct Dinic{ node arc[MAXE]; int vn, en, head[MAXV]; //vn点个数(包括源点汇点),en边个数 int cur[MAXV]; //当前弧 int q[MAXV]; //bfs建层次图时的队列 int path[MAXE], top; //存dfs当前最短路径的栈 int dep[MAXV]; //各节点层次 void init(int n){ vn = n; en = 0; mem(head, -1); } void insert_flow(int u, int v, int flow){ arc[en].u = u; arc[en].v = v; arc[en].flow = flow; arc[en].opp = en + 1; arc[en].next = head[u]; head[u] = en ++; arc[en].u = v; arc[en].v = u; arc[en].flow = flow; //反向弧 arc[en].opp = en - 1; arc[en].next = head[v]; head[v] = en ++; } bool bfs(int s, int t){ mem(dep, -1); int lq = 0, rq = 1; dep[s] = 0; q[lq] = s; while(lq < rq){ int u = q[lq ++]; if (u == t){ return true; } for (int i = head[u]; i != -1; i = arc[i].next){ int v = arc[i].v; if (dep[v] == -1 && arc[i].flow > 0){ dep[v] = dep[u] + 1; q[rq ++] = v; } } } return false; } int solve(int s, int t){ int maxflow = 0; while(bfs(s, t)){ int i, j; for (i = 1; i <= vn; i ++) cur[i] = head[i]; for (i = s, top = 0;;){ if (i == t){ int mink; int minflow = 0x3fffffff; for (int k = 0; k < top; k ++) if (minflow > arc[path[k]].flow){ minflow = arc[path[k]].flow; mink = k; } for (int k = 0; k < top; k ++) arc[path[k]].flow -= minflow, arc[arc[path[k]].opp].flow += minflow; maxflow += minflow; top = mink; //arc[mink]这条边流量变为0, 则直接回溯到该边的起点即可(这条边将不再包含在增广路内). i = arc[path[top]].u; } for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){ int v = arc[j].v; if (arc[j].flow && dep[v] == dep[i] + 1) break; } if (j != -1){ path[top ++] = j; i = arc[j].v; } else{ if (top == 0) break; dep[i] = -1; i = arc[path[-- top]].u; } } } return maxflow; } }dinic; bool vis[MAXV]; int num; void reach(int u){ vis[u] = 1; num ++; for (int i = dinic.head[u]; i != -1; i = dinic.arc[i].next){ int v = dinic.arc[i].v; if (vis[v] || dinic.arc[i].flow <= 0) continue; reach(v); } return ; } void work(int n, int A, int B){ num = 0; mem(vis, 0); reach(A); for (int i = 0; i < dinic.en; i += 2){ swap(dinic.arc[i].flow, dinic.arc[i^1].flow); } reach(B); if (num == n){ puts("UNIQUE"); } else{ puts("AMBIGUOUS"); } } int main(){ //freopen("test.in", "r", stdin); //freopen("test.out", "w", stdout); int n, m, A, B; while(scanf("%d %d %d %d", &n, &m, &A, &B), n){ dinic.init(n); for (int i = 0; i < m; i ++){ int u,v,w; scanf("%d %d %d", &u, &v, &w); dinic.insert_flow(u, v, w); } dinic.solve(A, B); work(n, A, B); } return 0; }
