-DP-树形DP-换根DP- [洛谷 P3047][USACO12FEB]附近的牛Nearby Cows

题目描述

Link

Solution

关于这道题我们可以先DP出每个节点的子树内距离为K的节点的权值和,然后我们再进行换根DP,处理出树上距离每个节点为K的节点点权和,最后把小于等于K的答案都加起来就行了。

Code

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 1000010
struct Edge {
int v, nx;
}e[MAXN << 1];
int head[MAXN], ecnt, n, m, x, y, k, dep[MAXN], dis[MAXN][22];
int dish[MAXN][22], ans[MAXN][22], tot[MAXN], disk[MAXN][22], c[MAXN][22];
void add(int f, int t) {
e[++ecnt] = (Edge) {t, head[f]};
head[f] = ecnt;
}
void dfs2(int u, int f, int p) {
for (int i = head[u]; i; i = e[i].nx) {
int to = e[i].v;
if (to == f) continue;
dis[u][p + 1] += dis[to][p];
dfs2(to, u, p);
}
}
void dfs3(int u, int f, int p) {
for (int i = head[u]; i; i = e[i].nx) {
int to = e[i].v;
if (to == f) continue;
disk[to][p] = disk[u][p - 1] - dis[to][p - 2] + dis[to][p];
dfs3(to, u, p);
}
}
void dp(int u, int f) {

}
int main() {
scanf("%d%d", &n, &k);
for (int i = 1; i < n; i++) {
scanf("%d%d", &x, &y);
add(x, y);
add(y, x);
}
for (int i = 1; i <= n; i++) {
scanf("%d", &dis[i][0]);
disk[i][0] = dis[i][0];
}
for (int i = 0; i < k; i++) {
dfs2(1, 1, i);
}
for (int i = 1; i <= k; i++) {
disk[1][i] = dis[1][i];
dfs3(1, 1, i);
}
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= k; j++) {
tot[i] = tot[i] + disk[i][j];
}
}
for (int i = 1; i <= n; i++) {
printf("%d\n", tot[i]);
}
}
文章作者: RiverFun
文章链接: https://stevebraveman.github.io/blog/2019/10/30/108/
版权声明: 本博客所有文章除特别声明外,均采用 CC BY-NC-SA 4.0 许可协议。转载请注明来自 RiverFun

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