Network（POJ3694+边双连通分量+LCA）

题目链接：http://poj.org/problem?id=3694

题目：

题意：给你一个n个点m条边的无向连通图，进行q次操作，每次操作在u和v之间加一条边，问每次操作之后“桥”的数量。

思路：先tarjan预处理出初始状态“桥”的数量cnt，并进行标记，对于每次操作，进行lca查询，将u和v之间的桥的数量num统计好，并消除标记，结果就是cnt-num。

代码实现如下：

 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 typedef pair<ll, ll> pll;
 typedef pair<ll, int> pli;
 typedef pair<int, ll> pil;;
 typedef pair<int, int> pii;
 typedef unsigned long long ull;

 #define lson i<<1
 #define rson i<<1|1
 #define bug printf("*********\n");
 #define FIN freopen("D://code//in.txt", "r", stdin);
 #define debug(x) cout<<"["<<x<<"]" <<endl;
 #define IO ios::sync_with_stdio(false),cin.tie(0);

 const double eps = 1e-;
 const int mod = ;
 const int maxn = 1e5 + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f;

 int n, m, q, u, v, tot, cnt, num;
 int head[maxn], dfn[maxn], low[maxn], vis[maxn], pre[maxn], c[maxn];

 struct edge {
     int v, next;
 }ed[maxn<<];

 void init() {
     tot = cnt = num = ;
     memset(c, , sizeof(c));
     memset(vis, , sizeof(vis));
     memset(dfn, , sizeof(dfn));
     memset(low, , sizeof(low));
     memset(head, -, sizeof(head));
     for(int i = ; i < maxn; i++) {
         pre[i] = i;
     }
 }

 void addedge(int u, int v) {
     ed[tot].v = v;
     ed[tot].next = head[u];
     head[u] = tot++;
     ed[tot].v = u;
     ed[tot].next = head[v];
     head[v] = tot++;
 }

 void tarjan(int x, int fa) {
     dfn[x] = low[x] = ++num;
     c[x] = ;
     for(int i = head[x]; ~i; i = ed[i].next) {
         int v = ed[i].v;
         if(!c[v]) {
             tarjan(v, x);
             pre[v] = x;
             low[x] = min(low[x], low[v]);
             if(low[v] > dfn[x]) {
                 vis[v] = ;
                 cnt++;
             }
         } else if(c[v] ==  && v != fa) {
             low[x] = min(low[x], dfn[v]);
         }
     }
     c[x] = ;
 }

 int cut(int x, int y) {
     int cnt = ;
     while(dfn[x] > dfn[y]) {
         if(vis[x]) {
             cnt++;
             vis[x] = ;
         }
         x = pre[x];
     }
     while(dfn[y] > dfn[x]) {
         if(vis[y]) {
             cnt++;
             vis[y] = ;
         }
         y = pre[y];
     }
     while(x != y) {
         if(vis[x]) {
             cnt++;
             vis[x] = ;
         }
         if(vis[y]) {
             cnt++;
             vis[y] = ;
         }
         x = pre[x];
         y = pre[y];
     }
     return cnt;
 }

 int main() {
     //FIN;
     int icase = ;
     while(~scanf("%d%d", &n, &m)) {
         if(n ==  && m == ) break;
         init();
         for(int i = ; i <= m; i++) {
             scanf("%d%d", &u, &v);
             addedge(u, v);
         }
         printf("Case %d:\n", ++icase);
         tarjan(, );
         scanf("%d", &q);
         while(q--) {
             scanf("%d%d", &u, &v);
             cnt -= cut(u, v);
             printf("%d\n", cnt);
         }
         printf("\n");
     }
     return ;
 }