POJ 3694 Network 无向图双联通+LCA

一开始题目没看清楚，以为是增加那条边后还有多少桥，所以就当做是无向图tarjan缩点后建树，然后求u，v的最近公共祖先，一直wa。

后来再看题目后才发现边放上去后不会拿下来了，即增加i条边后桥的数量。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int maxn = 100100;
const int maxm =  200100;

struct node{
    int v,next;
}edge[maxm*2];
int head[maxn],low[maxn],dfn[maxn],fa[maxn],stack[maxn],in[maxn],vis[maxm*2];
int res[maxn][2],depth[maxn],father[maxn];
int n,m,id,clock,top,total,num,q;
void add_edge(int u,int v){
    edge[id].v = v;edge[id].next = head[u];head[u] = id++;
    edge[id].v = u;edge[id].next = head[v];head[v] = id++;
}
void init(){
	memset(head,-1,sizeof(head));
    memset(dfn,0,sizeof(dfn));
	memset(fa,0,sizeof(fa));
    memset(vis,0,sizeof(vis));
    id = total = top = clock = num = 0;
    int u,v;
    while( m-- ){
        scanf("%d%d",&u,&v);
        add_edge(u,v);
    }
}
void tarjan(int u){
    low[u] = dfn[u] = ++clock;
    stack[top++] = u;in[u] = 1;
    for(int id = head[u]; id != -1; id = edge[id].next){
        if( vis[id] )continue;//如果边以及走过，则不能再走了
        vis[id] = vis[id^1] = 1;//标记边已经走过
        int v = edge[id].v;
        if(!dfn[v]){
            tarjan(v);
            if( dfn[u] < low[v])
            {//将桥两端的顶点保存下来
                res[total][0] = u;
                res[total++][1] = v;
            }
            low[u] = low[v] < low[u] ? low[v] : low[u];
        }
        else if(in[v] && low[u] > dfn[v])low[u] = dfn[v];
    }
    if( dfn[u] == low[u]){
         ++num;//缩点
        do{
            int v = stack[--top];
            fa[v] = num;
            in[v] = 0;
        }while( u != stack[top]);
    }
}

void dfs(int u,int dep){//确定个顶点的深度，以及其父亲节点
    depth[u] = dep;
    vis[u] = 1;
    for(int id= head[u]; id != -1; id = edge[id].next){
        int v = edge[id].v;
        if(vis[v])continue;
        father[v] = u;
        dfs(v,dep+1);
    }
}
int find(int x,int y){//递归找x，y的最近公共祖先
    if(x == y)return x;
   if(depth[x] > depth[y] )
   {
        if(!vis[x])total --;//当x所连的点为桥时，桥的数量减1
        vis[x] = 1;
        return find(father[x],y);
   }
   if(!vis[y])total --;
        vis[y] = 1;
    return find(x,father[y]);
}
void solve(){
	int i;
    tarjan(1);
    int u,v;
	memset(head,-1,sizeof(head));
	id= 0;
    for( i = 0 ; i < total; i++){//建树
        u = fa[res[i][0]], v = fa[res[i][1]];
        add_edge(u,v);
    }
    memset(vis,0,sizeof(vis));
    father[1] = 1;
    dfs(1,1);
}
int main(){
    int u,v;
  // freopen("in.txt","r",stdin);
    int cas = 1;
    while(~scanf("%d%d",&n,&m),n&&m){
        init();
        solve();
		scanf("%d",&q);
        printf("Case %d:
",cas++);
        memset(vis,0,sizeof(vis));
        for(int i = 1; i <= q; i ++)
        {
			scanf("%d%d",&u,&v);
			u = fa[u];v = fa[v];
			find(u,v);
			printf("%d
",total);
        }
        puts("");
    }
    return 0;
}