uva 10765 Doves and Bombs（割顶）

来源：互联网发布：编发教学软件编辑：程序博客网时间：2024/06/05 05:11

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;}

0 0