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题目链接：http://poj.org/problem?id=3694
题目大意：给定一个图，每次添加一条边（可能有重边）。输出每次添加后桥的
数目。由于添加边的次数比较多，添加一次Tarjin一次明显会超时。后来查到了
LCA算法，利用保存的子节点与最近父节点的关系进行计算的。第一次Tarjin后将
桥和其所在的父子节点关系保存下来，之后的m次添加边只需要在LCA中判断添加
的边是否为桥，若为桥则将此桥标记抹去，并将桥数减一，否则无影响。
附AC代码：
#include <stdio.h> #include <algorithm> #include <string.h> using namespace std; struct ad { int to, next, used; }edge[400005]; int head[100005], edge_num, dfn[100005], low[100005], bridge_num, Time, father[100005]; bool isbridge[100005]; void Add(int x, int y) { edge[edge_num].to = y; edge[edge_num].next = head[x]; edge[edge_num].used = 0; head[x] = edge_num++; } void Init() { memset(head, -1, sizeof(head)); memset(father, 0, sizeof(father)); memset(low, 0, sizeof(low)); memset(dfn, 0, sizeof(dfn)); memset(isbridge, false, sizeof(isbridge)); edge_num = Time = bridge_num = 0; } void Tarjin(int u, int fa) { father[u] = fa; low[u] = dfn[u] = ++Time; int i, v; for(i=head[u]; i!=-1; i=edge[i].next) { if(edge[i].used==0) { edge[i].used = edge[i^1].used = 1; v = edge[i].to; if(!dfn[v]) { Tarjin(v, u); low[u] = min(low[u], low[v]); if(low[v]>dfn[u]) { bridge_num++; isbridge[v] = true; } } else if(v!=fa) low[u] = min(low[u], dfn[v]); } } } void LCA(int u, int v) { if(dfn[u]>dfn[v]) swap(u, v); while(dfn[v]>dfn[u]) { if(isbridge[v])bridge_num--; isbridge[v] = false; v = father[v]; } while(u!=v) { if(isbridge[u])bridge_num--; if(isbridge[v])bridge_num--; isbridge[u] = isbridge[v] = false; u = father[u]; v = father[v]; } } int main() { int n, m, icase = 1; while(scanf("%d %d", &n, &m), n+m) { int a, b; Init(); for(int i=1; i<=m; i++) { scanf("%d %d", &a, &b); Add(a, b); Add(b, a); } Tarjin(1, -1); int Q; scanf("%d", &Q); printf("Case %d:\n", icase++); while(Q--) { scanf("%d %d", &a, &b); LCA(a, b); printf("%d\n", bridge_num); } } return 0; }
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原文地址：http://www.cnblogs.com/zznulw/p/5785260.html
