HDU - 4118 Holiday's Accommodation

2
4
1 2 3
2 3 2
4 3 2
6
1 2 3
2 3 4
2 4 1
4 5 8
5 6 5

Case #1: 18
Case #2: 62

题意：一颗树。相应1-n的结点，如今要求是每一个结点原本的人都走到不一样的结点去，每一个人都有路程，求总的最大路程

思路：对于每条边我们能想到的是：这条边左边的结点都跑到右边去，右边的 结点都跑到左边去。所以每条边。都会被走min{left[num], sum-left[nu,]}*2次，依据这个原则我们进行树形DP。可是这题递归的写法会跪。所以仅仅能手动DFS

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 200010;

struct Node {
	int to, next;
	int len;
} edge[maxn<<1];
int head[maxn], num[maxn], cnt, n;
int sta[maxn], vis[maxn];
ll ans;

void init() {
	cnt = 0;
	memset(head, -1, sizeof(head));
	memset(num, 0, sizeof(num));
}

void add(int u, int v, int len) {
	edge[cnt].to = v;
	edge[cnt].len = len;
	edge[cnt].next = head[u];
	head[u] = cnt++;

	edge[cnt].to = u;
	edge[cnt].len = len;
	edge[cnt].next = head[v];
	head[v] = cnt++;
}

void dfs(int u) {
	memset(vis, 0, sizeof(vis));
	int top = 0;
	sta[top++] = u;
	vis[u] = 1;
	while (top > 0) {
		int flag = 1;
		int cur = sta[top-1];
		for (int i = head[cur]; i != -1; i = edge[i].next) {
			int v = edge[i].to;
			if (vis[v]) continue;
			flag = 0;
			sta[top++] = v;
			vis[v] = 1;
		}

		if (flag == 0) continue;
		top--;

		for (int i = head[cur]; i != -1; i = edge[i].next) {
			int v = edge[i].to;
			if (num[v] != 0) {
				num[cur] += num[v];
				ans += (ll) edge[i].len * min(num[v], n - num[v]);
			}
		}
		num[cur]++;
	}
}

int main() {
	int t, cas = 1;
	int u, v, w;
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		init();
		ans = 0;
		for (int i = 1; i < n; i++) {
			scanf("%d%d%d", &u, &v, &w);
			add(u, v, w);
		}
		dfs(1);
		printf("Case #%d: %I64d\n", cas++, ans*2);
	}
	return 0;
}