POJ 3694 Network（并查集缩点 + 朴素的LCA + 无向图求桥）题解

题意：给你一个无向图，有q次操作，每次连接两个点，问你每次操作后有几个桥

思路：我们先用tarjan求出所有的桥，同时我们可以用并查集缩点，fa表示缩点后的编号，还要记录每个节点父节点pre。我们知道，缩点后形成一棵树，所有边都是桥，连接两点必会成环，环上任意边都不是桥。所以连点后，我们把两个点一步一步往上走，如果往上走之后发现fa不一样，说明走过了一条桥，那么合并fa，桥数量-1。

代码：

#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
typedef long long ll;
const int maxn = 100000 + 10;
const int seed = 131;
const ll MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
using namespace std;
int head[maxn], dfn[maxn], low[maxn], vis[maxn], fa[maxn], pre[maxn], tot, index, ans;
struct Edge{
    int to, next;
}e[maxn << 2];
void addEdge(int u, int v){
    e[tot].to = v;
    e[tot].next = head[u];
    head[u] = tot++;
}
int find(int x){
    return fa[x] == x? x : fa[x] = find(fa[x]);
}
void Union(int u, int v){
    int fx = find(u);
    int fy = find(v);
    if(fx != fy){
        fa[fx] = fy;
    }
}
void tarjan(int u, int Fa){
    vis[u] = 1;
    dfn[u] = low[u] = ++index;
    pre[u] = Fa;
    for(int i = head[u]; i != -1; i = e[i].next){
        int v = e[i].to;
        if(!dfn[v]){
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
            if(low[v] > dfn[u]){
                ans++;
            }
            else{
                Union(u, v);
            }
        }
        else if(v != Fa){
            low[u] = min(low[u], dfn[v]);
        }
    }
}
void Move(int x){
    int fx = find(x);
    int fy = find(pre[x]);
    if(fx != fy){
        fa[fx] = fy;
        ans--;
    }
}
void LCA(int u, int v){
    while(dfn[u] > dfn[v]){
        Move(u);
        u = pre[u];
    }
    while(dfn[v] > dfn[u]){
        Move(v);
        v = pre[v];
    }
    while(u != v){
        Move(u);
        Move(v);
        u = pre[u];
        v = pre[v];
    }
}
void init(){
    tot = index = ans = 0;
    memset(head, -1, sizeof(head));
    memset(vis, 0, sizeof(vis));
    memset(dfn, 0, sizeof(dfn));
}
int main(){
    int n, m, Case = 1;
    while(scanf("%d%d", &n, &m) && n + m){
        init();
        int u, v;
        for(int i = 0; i < m; i++){
            scanf("%d%d", &u, &v);
            addEdge(u, v);
            addEdge(v, u);
        }
        for(int i = 0; i <= n; i++)
            fa[i] = i;
        tarjan(1, 1);
        int q;
        scanf("%d", &q);
        printf("Case %d:
", Case++);
        while(q--){
            scanf("%d%d", &u, &v);
            if(find(u) != find(v)){
                LCA(u, v);
            }
            printf("%d
", ans);
        }
        printf("
");
    }
    return 0;
}