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
聚客AI7 小时前
🙋‍♀️Transformer训练与推理全流程:从输入处理到输出生成
人工智能·算法·llm
大怪v9 小时前
前端:人工智能?我也会啊!来个花活,😎😎😎“自动驾驶”整起!
前端·javascript·算法
惯导马工11 小时前
【论文导读】ORB-SLAM3:An Accurate Open-Source Library for Visual, Visual-Inertial and
深度学习·算法
骑自行车的码农13 小时前
【React用到的一些算法】游标和栈
算法·react.js
博笙困了13 小时前
AcWing学习——双指针算法
c++·算法
moonlifesudo14 小时前
322:零钱兑换(三种方法)
算法
NAGNIP1 天前
大模型框架性能优化策略:延迟、吞吐量与成本权衡
算法
美团技术团队1 天前
LongCat-Flash:如何使用 SGLang 部署美团 Agentic 模型
人工智能·算法
Fanxt_Ja2 天前
【LeetCode】算法详解#15 ---环形链表II
数据结构·算法·leetcode·链表
侃侃_天下2 天前
最终的信号类
开发语言·c++·算法