poj3694(lca + tarjan求桥模板)

题目链接: http://poj.org/problem?id=3694

题意: 给出一个 n 个节点 m 条边的图, 然后有 q 组形如 x, y 的询问, 在前面的基础上连接边 x, y, 输出当前图中有多少桥 .

思路: http://www.cnblogs.com/scau20110726/archive/2013/05/29/3106073.html

代码:

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 using namespace std;

 const int MAXN = 1e5 + ;

 struct node{
     int v, next, use;
 }edge[MAXN << ];

 bool bridge[MAXN];
 int low[MAXN], dfn[MAXN], vis[MAXN];
 int head[MAXN], pre[MAXN], ip, sol, count;

 void init(void){
     memset(head, -, sizeof(head));
     memset(vis, false, sizeof(vis));
     memset(bridge, false, sizeof(bridge));
     count = sol = ip = ;
 }

 void addedge(int u, int v){
     edge[ip].v = v;
     edge[ip].use = ;
     edge[ip].next = head[u];
     head[u] = ip++;
 }

 void tarjan(int u){
     vis[u] = ;
     dfn[u] = low[u] = count++;
     for(int i = head[u]; i != -; i = edge[i].next){
         if(!edge[i].use){
             edge[i].use = edge[i ^ ].use = ;
             int v = edge[i].v;
             if(!vis[v]){
                 pre[v] = u;
                 tarjan(v);
                 low[u] = min(low[u], low[v]);
                 if(dfn[u] < low[v]){
                     sol++;
                     bridge[v] = true;
                 }
             }else if(vis[v] == ){
                 low[u] = min(low[u], dfn[v]);
             }
         }
     }
     vis[u] = ;
 }

 void LCA(int u, int v){
     if(dfn[u] > dfn[v]) swap(u, v);
     while(dfn[v] > dfn[u]){//判断一下u,v是否在同一条树枝上
         if(bridge[v]) sol--;
         bridge[v] = false;
         v = pre[v];
     }
     while(u != v){//找lca
         if(bridge[u]) sol--;
         if(bridge[v]) sol--;
         bridge[u] = bridge[v] = false;
         u = pre[u];
         v = pre[v];
     }
 }

 int main(void){
     int n, m, q, x, y, cas = ;
     while(~scanf("%d%d", &n, &m)){
         if(!n && !m) break;
         init();
         for(int i = ; i < m; i++){
             scanf("%d%d", &x, &y);
             addedge(x, y);
             addedge(y, x);
         }
         pre[] = ;
         tarjan();
         printf("Case %d:\n", cas++);
         scanf("%d", &q);
         while(q--){
             scanf("%d%d", &x, &y);
             LCA(x, y);
             printf("%d\n", sol);
         }
         puts("");
     }
     return ;
 }