uva 10765 Doves and Bombs（割顶）

标签：图论   割顶   

题意:给定一个n个点的连通的无向图，一个点的“鸽子值”定义为将它从图中删去后连通块的个数。求每个点的“鸽子值”。

思路dfs检查每个点是否为割顶，并标记除去该点后有多少个连通分量

#include<cstdio>  
#include<cstring>  
#include<cmath>  
#include<cstdlib>  
#include<iostream>  
#include<algorithm>  
#include<vector>  
#include<map>  
#include<queue>  
#include<stack> 
#include<string>
#include<map> 
#include<set>
#define eps 1e-6 
#define LL long long  
using namespace std;  

const int maxn = 10000 + 100;
const int INF = 0x3f3f3f3f;
int n, m;
vector<int> G[maxn];
int val[maxn], node[maxn]; //node数组记录结点id间接排序 
bool cmp(int x, int y) {
	return val[x] == val[y] ? x < y : val[x] > val[y];
}

int pre[maxn], dfs_clock;
int dfs(int u, int fa) { //u在dfs树中的父节点为fa 
	int lowu = pre[u] = ++dfs_clock;
	int child = 0;        //子节点个数 
	for(int i = 0; i < G[u].size(); i++) {
		int v = G[u][i];
		if(!pre[v]) {   //没有访问过v 
			child++;
			int lowv = dfs(v, u);
			lowu = min(lowu, lowv);    //用后代的low函数更新u的low函数 
			if(lowv >= pre[u]) {
				val[u]++;
			}
				
		}
		else if(pre[v] < pre[u] && v != fa)  lowu = min(lowu, pre[v]);  //用反向边更新u的low函数 
	} 
	if(fa < 0 && child == 1) val[u] = 1;
	return lowu; 
} 


void init() {
	dfs_clock = 0;
	memset(pre, 0, sizeof(pre));
	for(int i = 0; i < n; i++) {
		G[i].clear();
		node[i] = i;
		val[i] = 1;
	}
	int x, y;
	while(scanf("%d%d", &x, &y) == 2 && x >= 0) {
		G[x].push_back(y);
		G[y].push_back(x);
	}
}

void solve() {
	dfs(0, -1);
	sort(node, node+n, cmp);
	for(int i = 0; i < m; i++) cout << node[i] << " " << val[node[i]] << endl; 
	cout << endl;
}

int main() {
	//freopen("input.txt", "r", stdin);
	while(scanf("%d%d", &n, &m) == 2 && n) {
		init();
		solve();
	}
	return 0;
}

uva 10765 Doves and Bombs（割顶）

标签：图论   割顶   

原文地址：http://blog.csdn.net/u014664226/article/details/46622001