K. Kingdom‘s Power,树形dp

Problem - K - Codeforces

time limit per test

2.0 s

memory limit per test

512 megabytes

input

standard input

output

standard output

Alex is a professional computer game player.

These days, Alex is playing a war strategy game. The kingdoms in the world form a rooted tree. Alex's kingdom 1 is the root of the tree. With his great diplomacy and leadership, his kingdom becomes the greatest empire in the world. Now, it's time to conquer the world!

Alex has almost infinite armies, and all of them are located at 1 initially. Every week, he can command one of his armies to move one step to an adjacent kingdom. If an army reaches a kingdom, that kingdom will be captured by Alex instantly.

Alex would like to capture all the kingdoms as soon as possible. Can you help him?

Input

The first line of the input gives the number of test cases, T (1≤T≤105). T test cases follow.

For each test case, the first line contains an integer n (1≤n≤106), where n is the number of kingdoms in the world.

The next line contains (n−1)integers f2,f3,⋯,fn (1≤fi<i), representing there is a road between fiand i.

The sum of n in all test cases doesn't exceed 5×106.

Output

For each test case, output one line containing "Case #x: y", where x is the test case number (starting from 1), and y is the minimum time to conquer the world.

Example

input

Copy

复制代码
2
3
1 1
6
1 2 3 4 4

output

Copy

复制代码
Case #1: 2
Case #2: 6

解析:

题意:

1号节点为根节点,根节点的军队数量是无线的,从根节点开始派兵,一次走一步,走遍所有的节点,需要多少步。

性质:若一棵子树都多个子树,这先走深度小的子树再走深度深的子树,是最优的;同时还要考虑是从一棵子树返回来走向另一棵子树还是从根节点派兵走

所以先dfs一次找出所有的节点的deep,再dfs找出最终答案;

具体见代码:(这道树形dp感觉跟dp没什么关系)

cpp 复制代码
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
using namespace std;
typedef long long LL;
const int N = 1e6 + 5;
int n;
vector<pair<int, int>>G[N];
int deep[N];
LL ans;
int d = 0, flg = 0;

int dfs(int x,int fa) {
	deep[x] = deep[fa] + 1;
	if (G[x].size() == 0)return 1;
	for (int i = 0; i < G[x].size(); i++) {
		int j = G[x][i].second;
		int& v = G[x][i].first;
		v=dfs(j, x);
	}
	sort(G[x].begin(), G[x].end());
	return G[x].back().first + 1;
}

void dfs1(int x,int fa) {
	if (G[x].size() == 0) {
		if (flg == 0)ans += deep[x];
		else {
			ans += min(deep[x], d);
		}
		flg = 1;
		d = 0;
		return ;
	}
	for (int i = 0; i < G[x].size(); i++) {
		int j = G[x][i].second;
		d++;
		dfs1(j, x);
		d++;
	}
}
int main() {
	int T,cnt=0;
	scanf("%d", &T);
	while (T--) {
		cnt++;
		scanf("%d", &n);
		for (int i = 0; i <= n; i++)
			G[i].clear();
		for (int i = 2,a; i <= n; i++) {
			scanf("%d", &a);
			G[a].push_back({ 0,i });
		}
		deep[0] = -1;
		dfs(1,0);
		flg = 0, d = 0, ans = 0;
		dfs1(1,0);
		printf("Case #%d: %lld\n",cnt, ans);
	}
	return 0;
}
相关推荐
九圣残炎23 分钟前
【从零开始的LeetCode-算法】1456. 定长子串中元音的最大数目
java·算法·leetcode
lulu_gh_yu29 分钟前
数据结构之排序补充
c语言·开发语言·数据结构·c++·学习·算法·排序算法
丫头,冲鸭!!!1 小时前
B树(B-Tree)和B+树(B+ Tree)
笔记·算法
Re.不晚1 小时前
Java入门15——抽象类
java·开发语言·学习·算法·intellij-idea
为什么这亚子2 小时前
九、Go语言快速入门之map
运维·开发语言·后端·算法·云原生·golang·云计算
2 小时前
开源竞争-数据驱动成长-11/05-大专生的思考
人工智能·笔记·学习·算法·机器学习
~yY…s<#>2 小时前
【刷题17】最小栈、栈的压入弹出、逆波兰表达式
c语言·数据结构·c++·算法·leetcode
幸运超级加倍~3 小时前
软件设计师-上午题-16 算法(4-5分)
笔记·算法
yannan201903133 小时前
【算法】(Python)动态规划
python·算法·动态规划
埃菲尔铁塔_CV算法3 小时前
人工智能图像算法:开启视觉新时代的钥匙
人工智能·算法