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题意:
有n个电脑1-n,m个连接,由于可能存在一些桥,如果这些桥出现了问题,那么会导致一些电脑之间无法连接,
所以建立链接Q次,每次链接a和b电脑,求链接ab后还存在几个桥;
如果说在ab之间建立连接, a和b的最近祖先(LCA),那么从a到LCA之间和b到LCA之间的桥都将不在是桥了
#include <iostream> #include <cstdlib> #include <cstdio> #include <algorithm> #include <vector> #include <queue> #include <cmath> #include <stack> #include <cstring> using namespace std; #define INF 0xfffffff #define N 101000 int low[N], dfn[N], f[N], bridge[N]; //bridge[i] 是表示i和f[i]之间的路是否是桥;值为1的时候是桥; vector<vector<int> > G; int n, Time, m, nbridge; void Init() { G.clear(); G.resize(n+1); Time = nbridge = 0; memset(low, 0, sizeof(low)); memset(dfn, 0, sizeof(dfn)); memset(f, 0, sizeof(f)); } void Tarjan(int u, int father) { f[u] = father; dfn[u] = low[u] = ++Time; int len = G[u].size(), v; for(int i=0; i<len; i++) { v = G[u][i]; if(!dfn[v]) { Tarjan(v, u); low[u] = min(low[u], low[v]); if(low[v] > dfn[u]) { bridge[v]++; nbridge++; } } else if(v!=father)//避免找回自己的父节点; { low[u] = min(low[u], dfn[v]); } } } void LCA(int a, int b) { if( a == b ) return ; if(dfn[a] < dfn[b]) { if(bridge[b]==1) { bridge[b] = 0; nbridge--; } LCA(a, f[b]); } else { if(bridge[a]==1) { bridge[a] = 0; nbridge--; } LCA(f[a], b); } } int main() { int t=0, a, b, Q; while(scanf("%d %d", &n, &m), n+m) { Init(); while(m--) { scanf("%d %d", &a, &b); G[a].push_back(b); G[b].push_back(a); } Tarjan(1,0); printf("Case %d:\n", ++t); scanf("%d", &Q); while(Q--) { scanf("%d%d", &a,&b); LCA(a,b); printf("%d\n",nbridge); } printf("\n"); } return 0; }
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原文地址:http://www.cnblogs.com/zhengguiping--9876/p/4709732.html