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:
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 6 const int MAXN = 1e5 + 10; 7 8 struct node{ 9 int v, next, use; 10 }edge[MAXN << 2]; 11 12 bool bridge[MAXN]; 13 int low[MAXN], dfn[MAXN], vis[MAXN]; 14 int head[MAXN], pre[MAXN], ip, sol, count; 15 16 void init(void){ 17 memset(head, -1, sizeof(head)); 18 memset(vis, false, sizeof(vis)); 19 memset(bridge, false, sizeof(bridge)); 20 count = sol = ip = 0; 21 } 22 23 void addedge(int u, int v){ 24 edge[ip].v = v; 25 edge[ip].use = 0; 26 edge[ip].next = head[u]; 27 head[u] = ip++; 28 } 29 30 void tarjan(int u){ 31 vis[u] = 1; 32 dfn[u] = low[u] = count++; 33 for(int i = head[u]; i != -1; i = edge[i].next){ 34 if(!edge[i].use){ 35 edge[i].use = edge[i ^ 1].use = 1; 36 int v = edge[i].v; 37 if(!vis[v]){ 38 pre[v] = u; 39 tarjan(v); 40 low[u] = min(low[u], low[v]); 41 if(dfn[u] < low[v]){ 42 sol++; 43 bridge[v] = true; 44 } 45 }else if(vis[v] == 1){ 46 low[u] = min(low[u], dfn[v]); 47 } 48 } 49 } 50 vis[u] = 2; 51 } 52 53 void LCA(int u, int v){ 54 if(dfn[u] > dfn[v]) swap(u, v); 55 while(dfn[v] > dfn[u]){//判断一下u,v是否在同一条树枝上 56 if(bridge[v]) sol--; 57 bridge[v] = false; 58 v = pre[v]; 59 } 60 while(u != v){//找lca 61 if(bridge[u]) sol--; 62 if(bridge[v]) sol--; 63 bridge[u] = bridge[v] = false; 64 u = pre[u]; 65 v = pre[v]; 66 } 67 } 68 69 int main(void){ 70 int n, m, q, x, y, cas = 1; 71 while(~scanf("%d%d", &n, &m)){ 72 if(!n && !m) break; 73 init(); 74 for(int i = 0; i < m; i++){ 75 scanf("%d%d", &x, &y); 76 addedge(x, y); 77 addedge(y, x); 78 } 79 pre[1] = 1; 80 tarjan(1); 81 printf("Case %d: ", cas++); 82 scanf("%d", &q); 83 while(q--){ 84 scanf("%d%d", &x, &y); 85 LCA(x, y); 86 printf("%d ", sol); 87 } 88 puts(""); 89 } 90 return 0; 91 }