POJ 3352 Road Construction POJ 3177 Redundant Paths（边双连通图 Tarjan+缩点）

ACM

题目地址： 
POJ 3352 Road Construction 
POJ 3177 Redundant Paths

题意： 
问要添加几条边才能使所给无向图图变成边双连通图。

分析： 
边连通度：使无向图G不连通的最少删边数量为其边连通度。 
边双连通图：边连通度大于1的无向图。

首先缩点，让图变成一个DAG。 
现在问题转化为：在树中至少添加多少条边能使图变为双连通图。 
若要使得任意一棵树，在增加若干条边后，变成一个双连通图，那么至少增加的边数 =（ 这棵树总度数为1的结点数 + 1 ）/ 2。 
其实就是让树的叶子节点都成对相连，可以在纸上画看看就知道了。

代码：

/**  Author:      illuz <iilluzen[at]gmail.com>*  Blog:        http://blog.csdn.net/hcbbt*  File:        3352.cpp*  Create Date: 2014-08-13 15:28:44*  Descripton:   */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <stack>using namespace std;#define repf(i,a,b) for(int i=(a);i<=(b);i++)typedef long long ll;const int N = 5010;vector<int> G[N];bool app[N][N], instack[N];int n, r, u, v;int dfn[N], low[N], dg[N], tclock;void tarjan(int u, int fa) {dfn[u] = low[u] = ++tclock;instack[u] = 1;int sz = G[u].size();repf (i, 0, sz - 1) {int v = G[u][i];if (v == fa)continue;if (!dfn[v]) {tarjan(v, u);low[u] = min(low[u], low[v]);} else if (instack[v]) {low[u] = min(low[u], dfn[v]);}}}void init(int n, int r) {memset(app, 0, sizeof(app));memset(low, 0, sizeof(low));memset(dfn, 0, sizeof(dfn));memset(instack, 0, sizeof(instack));memset(dg, 0, sizeof(dg));repf (i, 1, n) {G[i].clear();}repf (i, 0, r - 1) {scanf("%d%d", &u, &v);if (!app[u][v]) {G[u].push_back(v);G[v].push_back(u);app[u][v] = app[v][u] = 1;}}}void solve() {tarjan(1, -1);repf (u, 1, n) {int sz = G[u].size();repf (i, 0, sz - 1) {int v = G[u][i];if (low[u] != low[v])dg[low[u]]++;}}int cnt = 0;repf (i, 1, n) {if (dg[i] == 1) {cnt++;}}printf("%d\n", (cnt + 1) / 2);}int main() {while (~scanf("%d%d", &n, &r)) {init(n, r);solve();}return 0;}