zoj 3820 Building Fire Stations The 2014 ACM-ICPC Asia Mudanjiang Regional Contest bfs
2014-10-23 23:26
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传送门:2014牡丹江区域赛B题
给定一棵树,从整棵树中选出两个节点,每个节点会选择这两个节点中离自己更近的点作为目标点,求怎样选点能使得所有点目标点的距离的最大值最小,输出这个距离值,以及选为目标点的编号
分析可以得出,节点选在直径上时最优。
当只选一个点时,选直径上的中点时最优。
这里要选两个点,可以先求一次直径。然后在最长直径上从中间把整棵树分为两棵树,然后分别求这两棵树的直径,分别求出中点与距离值即可
给定一棵树,从整棵树中选出两个节点,每个节点会选择这两个节点中离自己更近的点作为目标点,求怎样选点能使得所有点目标点的距离的最大值最小,输出这个距离值,以及选为目标点的编号
分析可以得出,节点选在直径上时最优。
当只选一个点时,选直径上的中点时最优。
这里要选两个点,可以先求一次直径。然后在最长直径上从中间把整棵树分为两棵树,然后分别求这两棵树的直径,分别求出中点与距离值即可
/****************************************************** * File Name: 3820.cpp * Author: kojimai * Create Time: 2014年10月21日 星期二 14时28分52秒 ******************************************************/ #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<iostream> #include<queue> using namespace std; #define FFF 200005 int first[FFF],e,v[FFF*2],next[FFF*2],pre[FFF],level[FFF]; void addedge(int x,int y) { v[e] = y; next[e] = first[x]; first[x] = e++; v[e] = x; next[e] = first[y]; first[y] = e++; } queue<int> p; int bfs(int x) { int ret = x,len = 0; while(!p.empty()) p.pop(); level[x] = 0; pre[x] = 0; p.push(x); while(!p.empty()) { int now = p.front(); p.pop(); for(int k = first[now]; ~k; k=next[k]) { int va = v[k]; if(pre[va] == -1) { pre[va] = now; level[va] = level[now] + 1; if(len == level[now]) { len = level[now] + 1; ret = va; } p.push(va); } } } return ret; } int geta(int now,int le) { while(level[now] > le/2) now = pre[now]; return now; } int main() { int keng,n,x,y,ans; scanf("%d",&keng); while(keng--) { e = 0; memset(first,-1,sizeof(first)); memset(pre,-1,sizeof(first)); scanf("%d",&n); for(int i = 1;i < n;i++) { scanf("%d%d",&x,&y); addedge(x,y); } int tmp = bfs(1);//先任选一点找深度最深的点,该点必为直径的一个端点 memset(pre,-1,sizeof(pre)); int tmp1 = bfs(tmp);//以找到的直径端点为起点,找深度最深的叶子节点,即直径的另一个端点 int len = level[tmp1] + 1;//求得的直径长度 //cout<<" len = "<<len<<endl; int t1,t2,ans1,ans2; t2 = tmp1; while(level[pre[t2]]+1 > len/2) { t2 = pre[t2]; } t1 = pre[t2]; //找出直径上中间点的位置,并由找到的两点将整棵树分为两棵子树 //cout<<"t1 = "<<t1<<" t2 = "<<t2<<endl; memset(pre,-1,sizeof(pre)); pre[t2] = 0; tmp = bfs(t1); memset(pre,-1,sizeof(pre)); pre[t2] = 0; tmp1 = bfs(tmp);//求出第一棵子树的直径 //cout<<"len1 = "<<level[tmp1]+1<<endl; int ans = (level[tmp1] + 1)/2;//直径的一半即为该子树上的距离最大值 ans1 = geta(tmp1,level[tmp1]);//找出直径上的中点 tmp = bfs(t2); memset(pre,-1,sizeof(pre)); pre[t1] = 0; tmp1 = bfs(tmp);//求出第二棵子树的直径 //cout<<"len2 = "<<level[tmp1]+1<<endl; ans = max(ans,(level[tmp1] + 1)/2);//距离取较大值 ans2 = geta(tmp1,level[tmp1]);//第二棵子树的中点 cout<<ans<<' '<<ans1<<' '<<ans2<<endl; } return 0; }
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