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于 2016-08-02 21:42:45 发布

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分类专栏：树的直径

本文链接：https://blog.youkuaiyun.com/blue_hpu/article/details/52098219

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本文介绍了一个关于寻找在特定节点间旅行最短路径的问题。在一个由N个景点通过N-1条双向路径连接的大公园里，主人公希望从这些景点中选择K个进行游览，并通过巧妙的算法找出从任一入口进入，遍历这K个景点所需的最短总距离。

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Park Visit

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3271 Accepted Submission(s): 1463

Problem Description

Claire and her little friend, ykwd, are travelling in Shevchenko's Park! The park is beautiful - but large, indeed. N feature spots in the park are connected by exactly (N-1) undirected paths, and Claire is too tired to visit all of them. After consideration, she decides to visit only K spots among them. She takes out a map of the park, and luckily, finds that there're entrances at each feature spot! Claire wants to choose an entrance, and find a way of visit to minimize the distance she has to walk. For convenience, we can assume the length of all paths are 1.
Claire is too tired. Can you help her?

Input

An integer T(T≤20) will exist in the first line of input, indicating the number of test cases.
Each test case begins with two integers N and M(1≤N,M≤10⁵), which respectively denotes the number of nodes and queries.
The following (N-1) lines, each with a pair of integers (u,v), describe the tree edges.
The following M lines, each with an integer K(1≤K≤N), describe the queries.
The nodes are labeled from 1 to N.

Output

For each query, output the minimum walking distance, one per line.

Sample Input

Sample Output

1
4

就是给出点的关系，问走过n需要最少走几条边，一次回头路算一条边，所以n不大于最长路就是n-1，大于的话，根据神奇的数学定理就是，最长路有k条边，结果就是k-1+（n-k）*2

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define maxn 100010 
using namespace std;
int ans, num, ok, nod, cnt;
int dis[maxn], head[maxn];
bool vis[maxn];
struct node{
	int from, to, val, next;
}edge[maxn*2];
void add(int u, int v, int w){
	edge[num].from = u;
	edge[num].to = v;
	edge[num].val = w;
	edge[num].next = head[u];
	head[u] = num++;
}
void bfs(int t){
	memset(vis, false, sizeof(vis));
	memset(dis, 0, sizeof(dis));
	queue<int > Q;
	Q.push(t);
	vis[t] = true;
	ans = 0;
	while(!Q.empty()){
		int u = Q.front();
		Q.pop();
		for(int i = head[u]; i != -1; i = edge[i].next){
			int v = edge[i].to;
			if(!vis[v]){
				dis[v] = edge[i].val + dis[u];
				vis[v] = true;
				Q.push(v);
				if(ans < dis[v]){
					ans = dis[v];
					nod = v;
				}
			}
		}
	}
}
int main(){
	int m, n, a, b, c, t;
	scanf("%d", &t);
	while(t--){
		scanf("%d%d", &m, &n);
		memset(head, -1, sizeof(head));
		num = ok = 0;
		for(int i = 0; i < m-1; i++){
			scanf("%d%d", &a, &b);
			add(a, b, 1);
			add(b, a, 1);
		}
		bfs(1);
		bfs(nod);
		for(int i = 0; i < n; i++){
			scanf("%d", &c);
			if(c-1 <= ans){
				printf("%d\n", c-1);
			}
			else{
				printf("%d\n", ans + (c-(ans+1))*2);
			}
		}
	}
	return 0;
}