洛谷P1351 联合权值(树形dp)

题意

题目链接

Sol

一道很简单的树形dp，然而被我写的这么长

分别记录下距离为\(1/2\)的点数，权值和，最大值。以及相邻儿子之间的贡献。

树形dp一波。。

#include<bits/stdc++.h>
#define Fin(x) {freopen(x, "r", stdin);}
#define int long long 
using namespace std;
const int MAXN = 2e5 + 10, mod = 10007;
inline int read() {
	char c = getchar(); int x = 0, f = 1;
	while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * f;
}
int N, a[MAXN];
vector<int> v[MAXN];
int f[MAXN][3], sum[MAXN][3], num[MAXN][3],  dis[MAXN], down[MAXN], Son[MAXN];
int add(int x, int y) {
	if(x + y < 0) return x + y + mod;
	else return x + y >= mod ? x + y - mod : x + y;
}
int mul(int x, int y) {
	return 1ll * x * y % mod;
}
void dfs(int x, int fa) {
	dis[x] = 0;
	for(int i = 0, to; i < v[x].size(); i++) {
		if((to = v[x][i]) == fa) continue;
		dfs(to, x);
		dis[x] = add(dis[x], mul(sum[x][1], a[to]));
		if(!Son[x]) Son[x] = a[to];
		else down[x] = max(down[x], a[to] * Son[x]), Son[x] = max(Son[x], a[to]);
		f[x][1] = max(f[x][1], a[to]);
		f[x][2] = max(f[x][2], f[to][1]);
		sum[x][1] = add(sum[x][1], a[to]);
		sum[x][2] = add(sum[x][2], sum[to][1]);
		num[x][1]++;
		num[x][2] += num[to][1];
	}
}
signed main() {
	memset(f, -1, sizeof(f));
	N = read();
	for(int i = 1; i <= N - 1; i++) {
		int x = read(), y = read();
		v[x].push_back(y); v[y].push_back(x);
	}
	for(int i = 1; i <= N; i++) a[i] = read();
	dfs(1, 0);
	int mx = 0, s = 0;
	for(int i = 1; i <= N; i++) {
		if(~f[i][2]) mx = max(mx, a[i] * f[i][2]);
		if(num[i][2]) s = add(s, mul(a[i], sum[i][2]));
		if(num[i][1] > 1) s = add(s, dis[i]), mx = max(mx, down[i]);
	}
	cout << mx << " " << s * 2 % mod;
	return 0;
}
/*
*/