POJ 3694 Network（双连通分量）
http://poj.org/problem?id=3694
题意 ： N个点，M条边，构成一个连通图，将图中加入特定的一些边，每加一条边就输出加入这条边后图中剩的桥的个数。
思路 ：双连通分量，用LCA
#include <iostream> #include <vector> #include <queue> #include <algorithm> #include <stdio.h> #include <string.h> using namespace std; const int maxn = 100005 ; const int maxm = 555555 ; const int INF = 1000000000 ; struct Edge { int v, next; }map[maxm]; int low[maxn],dfn[maxn],index,vis[maxn] ; int e,n,m,a,b,head[maxn] ; int cnt,bridge[maxn],father[maxn] ; void Init() { e = 0 ; index = 0 ; cnt = 0 ; memset(vis,0,sizeof(vis)) ; memset(bridge,0,sizeof(bridge)) ; memset(dfn,0,sizeof(dfn)) ; memset(head,-1,sizeof(head)) ; for(int i = 1 ; i <= n ; i++) father[i] = i ; } void addedge(int v,int w) { map[e].v = w ; map[e].next = head[v] ; head[v] = e++ ; } void tarjan(int u) { vis[u] = 1 ; dfn[u] = low[u] = ++index ; for(int i = head[u] ; i+1 ; i = map[i].next) { int v = map[i].v ; if(!vis[v]) { father[v] = u ; tarjan(v) ; low[u] = min(low[u],low[v]) ; if(low[v] > dfn[u]) { cnt++ ; bridge[v] = 1 ; } } else if(vis[v] == 1 && v != father[u]) low[u] = min(low[u],dfn[v]) ; } vis[u] = 2 ; } void lca(int u,int v) { while(dfn[u] > dfn[v]) { if(bridge[u]) cnt--,bridge[u] = 0 ; u = father[u] ; } while(dfn[v] > dfn[u]) { if(bridge[v]) cnt--,bridge[v] = 0 ; v = father[v] ; } while(u != v) { if(bridge[u]) cnt--,bridge[u] = 0 ; if(bridge[v]) cnt--,bridge[v] = 0 ; u = father[u] ; v = father[v] ; } } int main() { int kase = 0 ; while(scanf("%d %d",&n,&m)!=EOF) { if(n == 0 && m == 0) break ; Init() ; while(m--) { scanf("%d %d",&a,&b) ; addedge(a,b) ; addedge(b,a) ; } tarjan(1) ; int k ; scanf("%d",&k) ; printf("Case %d:\n",++kase) ; while(k--) { scanf("%d %d",&a,&b) ; lca(a,b) ; printf("%d\n",cnt) ; } printf("\n") ; } return 0; }