POJ 3177 Redundant Paths （tarjan bcc）

题意:

无向图问最少添加多少条边使得图成为双连通的

分析:

显然先把图进行bcc缩点,添加新的边形成bcc

显然应该选择距离最远的两个叶子节点,这样可以构成一个很大的环(bcc),合并这个bcc上的点,重复这个过程,直到整个图都成为1个bcc

显然新添加的边的个数应该是(leaf+1)/2

坑:

图有重边,判断的时候只要去掉是自己父亲这条边就好了

代码:

//
//  Created by TaoSama on 2015-11-21
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 5e3 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int M = 2e4 + 10;

int n, m;

struct Edge {
int v, nxt;
} edge[M];
int head
, cnt;

void addEdge(int u, int v) {
edge[cnt] = (Edge) {v, head[u]};
head[u] = cnt++;
edge[cnt] = (Edge) {u, head[v]};
head[v] = cnt++;
}

int dfn
, low
, in
, id
, bcc, dfsNum;
int stk
, top;

void tarjan(int u, int f) {
dfn[u] = low[u] = ++dfsNum;
stk[++top] = u;
in[u] = true;
for(int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
if(i == (f ^ 1)) continue;
if(!dfn[v]) {
tarjan(v, i);
low[u] = min(low[u], low[v]);
} else if(in[v]) low[u] = min(low[u], dfn[v]);
}
if(low[u] == dfn[u]) {
++bcc;
while(true) {
int v = stk[top--];
in[v] = false;
id[v] = bcc;
if(v == u) break;
}
}
}

void init() {
bcc = dfsNum = 0;
memset(dfn, 0, sizeof dfn);
cnt = 0;
memset(head, -1, sizeof head);
}

int deg
;

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

while(scanf("%d%d", &n, &m) == 2) {
init();
while(m--) {
int u, v; scanf("%d%d", &u, &v);
addEdge(u, v);
}
for(int i = 1; i <= n; ++i) if(!dfn[i]) tarjan(i, -1);

memset(deg, 0, sizeof deg);
for(int i = 1; i <= n; ++i) {
int u = id[i];
for(int j = head[i]; ~j; j = edge[j].nxt) {
int v = id[edge[j].v];
if(u == v) continue;
++deg[u]; ++deg[v];
}
}
int leaf = 0;
//each vertex has been counted 2 times
for(int i = 1; i <= bcc; ++i) leaf += deg[i] == 2;
printf("%d\n", leaf + 1 >> 1);
}
return 0;
}