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:\n",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\n",total);
}
puts("");
}
return 0;
}

  

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