Hihocoder 1063 缩地

树形dp

涉及不重复背包组合求最小

从边长分段看不好入手

因为点数只有100点值<=2，总值<=200

可以对每个点的每个值进行dp

这里最后不回来肯定优于全回来

然后由于要分为回来和不回来两种情况要分别dp，因为不回来会要用到回来的

不回来的可以按不回来的最小+去掉不回来那个子节点的回来的最小进行dp

另外注意下上下限和初始值

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
const int N = 205;
int vec[N];
struct Edge{
	int v, len, nxt;
}edge[N];
int cnt;
int head[N];
int dp[2][N][N];
void addedge(int u, int v, int value) {
	edge[cnt] = Edge{ v,value,head[u] };
	head[u] = cnt++;
	edge[cnt] = Edge{ u,value,head[v] };
	head[v] = cnt++;
}
void dfs(int u,int p) {
	dp[0][u][vec[u]] = 0;
	dp[1][u][vec[u]] = 0;
	for (int t = head[u]; t != -1; t = edge[t].nxt) {
		Edge e = edge[t];
		if (e.v == p)
			continue;
		dfs(e.v, u);	
		int v = e.v;
		for (int i = 200; i >= 0; i--)
		{
			for (int j = 0; j <= i; j++) {
				dp[0][u][i] = min(dp[0][u][i],dp[0][u][j] + dp[0][v][i-j]+2*e.len);
			}
		}
	}
	for (int t = head[u]; t != -1; t = edge[t].nxt) {
		Edge e = edge[t];
		int v = e.v;
		if (v == p)
			continue;
		int sum[N];
		for (int i = 0; i <= 202; i++)
			sum[i] = 1e9;
		sum[vec[u]] = 0;
		for (int t2 = head[u]; t2 != -1;t2=edge[t2].nxt) {
			Edge e2 = edge[t2];
			int v2 = e2.v;
			if (v2==p || v2 == v)
				continue;
			for (int i = 200; i >= 0; i--){
				for (int j = 0; j <= i; j++) {
					sum[i] = min(sum[i], sum[j] + dp[0][v2][i - j]+2*e2.len);
				}
			}
		}
		for (int i = 200; i >= 0; i--) {
			for (int j = 0; j <= i; j++) {
				dp[1][u][i] = min(dp[1][u][i], dp[1][v][j] + sum[i - j]+e.len);
			}
		}
	}
}
int main() {
	int n;
	cin >> n;
	cnt = 1;
	for (int i = 1; i <= n; i++) {
		cin >> vec[i];
	}
	memset(head, -1, sizeof head);
	for (int i = 0; i < 2; i++)
		for (int j = 0; j <= 200; j++)
			for (int k = 0; k <= 200; k++)
				dp[i][j][k] = 1e9;
	for (int i = 0; i < n - 1; i++) {
		int sv, ev, w;
		cin >> sv >> ev >> w;
		addedge(sv,ev,w);
	}
	dfs(1, 0);
	int m;
	cin >> m;
	while(m--){
		int x;
		cin >> x;
		for (int i = 200; i >=0; i--)
			if (dp[1][1][i] <= x)
			{
				printf("%d\n", i);
				break;
			}
	}
	return 0;
}