D. Police Stations----多起点bfs
来源:互联网 发布:虚拟机软件有哪些 知乎 编辑:程序博客网 时间:2024/06/06 09:40
Inzane finally found Zane with a lot of money to spare, so they together decided to establish a country of their own.
Ruling a country is not an easy job. Thieves and terrorists are always ready to ruin the country's peace. To fight back, Zane and Inzane have enacted a very effective law: from each city it must be possible to reach a police station by traveling at most d kilometers along the roads.
There are n cities in the country, numbered from 1 to n, connected only by exactly n - 1 roads. All roads are 1 kilometer long. It is initially possible to travel from a city to any other city using these roads. The country also has k police stations located in some cities. In particular, the city's structure satisfies the requirement enforced by the previously mentioned law. Also note that there can be multiple police stations in one city.
However, Zane feels like having as many as n - 1 roads is unnecessary. The country is having financial issues, so it wants to minimize the road maintenance cost by shutting down as many roads as possible.
Help Zane find the maximum number of roads that can be shut down without breaking the law. Also, help him determine such roads.
The first line contains three integers n, k, and d (2 ≤ n ≤ 3·105, 1 ≤ k ≤ 3·105, 0 ≤ d ≤ n - 1) — the number of cities, the number of police stations, and the distance limitation in kilometers, respectively.
The second line contains k integers p1, p2, ..., pk (1 ≤ pi ≤ n) — each denoting the city each police station is located in.
The i-th of the following n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the cities directly connected by the road with index i.
It is guaranteed that it is possible to travel from one city to any other city using only the roads. Also, it is possible from any city to reach a police station within d kilometers.
In the first line, print one integer s that denotes the maximum number of roads that can be shut down.
In the second line, print s distinct integers, the indices of such roads, in any order.
If there are multiple answers, print any of them.
6 2 41 61 22 33 44 55 6
15
6 3 21 5 61 21 31 41 55 6
24 5
In the first sample, if you shut down road 5, all cities can still reach a police station within k = 4 kilometers.
In the second sample, although this is the only largest valid set of roads that can be shut down, you can print either 4 5 or 5 4 in the second line.
题目链接:http://codeforces.com/contest/796/problem/D
题目的意思就是有一些特殊点,要求所有点距离特殊点的距离不能超过k,问你最多能去掉几条边。
我对于这个题除了暴力并没有任何的思路,然后无奈的我上网看了题解,多起点bfs。
给出大神的思路:
根据题意可以知道只要不删除边就必定满足条件,那么只要暴力把全部特殊点放进队列广搜一边就好了。若当前边走过就返回,若当前边没走过且下一个点走过那么这条边是可以删除的。
代码:
#include <cstdio>#include <cstring>#include <iostream>#include <queue>#define LL long longusing namespace std;const int maxn=300005;int vis[maxn];queue<int >Q;struct node{ int v,next;}p[maxn<<1];int head[maxn],len,vis1[maxn],vis2[maxn],ans[maxn],sum;void connect(int u,int v){ p[len].v=v; p[len].next=head[u]; head[u]=len++;}void bfs(){ while(!Q.empty()){ int now=Q.front(); Q.pop(); for(int i=head[now];~i;i=p[i].next){ if(vis2[i/2+1]) continue; vis2[i/2+1]=1; if(vis1[p[i].v]) ans[sum++]=i/2+1; else{ vis1[p[i].v]=1; Q.push(p[i].v); } } }}int main(){ int n,m,k; memset(head,-1,sizeof(head)); scanf("%d%d%d",&n,&m,&k); for(int i=0;i<m;i++){ int x; scanf("%d",&x); vis1[x]=1; Q.push(x); } for(int i=0;i<n-1;i++){ int u,v; scanf("%d%d",&u,&v); connect(u,v); connect(v,u); } bfs(); printf("%d\n",sum); for(int i=0;i<sum;i++){ printf(i==0?"%d":" %d",ans[i]); } cout<<endl; return 0;}
- D. Police Stations----多起点bfs
- Codeforces Round #408 (Div. 2)-D. Police Stations-多起点bfs
- Codeforces 796D Police Stations 构造+BFS
- 796D Police Stations
- Codeforces Round #408 (Div. 2) D.Police Stations【Bfs+思维】
- Codeforces Round #408 (Div. 2) -- D. Police Stations(bfs)
- Codeforces 796D Police Stations (bfs+思维)
- Codeforces Round #408 (Div. 2) D.Police Stations【Bfs+思维】
- Codeforces Round #408 (Div. 2) D. Police Stations 最短路、BFS
- Codeforces Round #408 D. Police Stations
- Police Stations
- Codeforces Round #408 (Div. 2) D. Police Stations
- Codeforces Round #408 (Div. 2)-D. Police Stations
- codeforces796D Police Stations
- Fire Game (多起点BFS)
- FZU 2150 Fire Game 多起点BFS
- FZU 2150 Fire Game 多起点BFS
- UVA 11624 Fire(bfs + 多起点)
- 抓包工具fiddler
- 中特基础整理
- 数据结构课下复习3
- 机器学习练习一:简单线性回归
- IOS学习(1)— IOS默认工程结构
- D. Police Stations----多起点bfs
- windows部署tomcat项目(2)
- [Linux基础]Linux基础知识入门及常见命令.
- 小强ROS机器人教程(10)___使用kinect进行自主移动避障
- 文章投稿推荐
- windows下node.js环境配置
- Java数组和foreach遍历循环
- BZOJ 1012: [JSOI2008]最大数maxnumber
- 好书推荐:Python网络数据采集