[USACO10MAR]Great Cow Gathering G（换根dp）

标签：getc   oid   pac   long   problem   node   char   i++   print   

[USACO10MAR]Great Cow Gathering G

换根dp模板题。

同时记录\(sz[u]\)代表\(u\)的子树内有多少奶牛，那转移时即为\(dp[u]=dp[v]+sz[v] \times val(u,v)\)。

注意开long long。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 100010;
const ll inf = 0x7f7f7f7f7f7f7f7f;
template <typename T> void read(T &x) {
	T f = 1;
	char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == ‘-‘) f = -1;
	for (x = 0; isdigit(ch); ch = getchar()) x = x * 10 + ch - ‘0‘;
	x *= f;
}
template <typename T> void cmin(T &x, T y) {if (y < x) x = y;}
struct node{
	int pre, to;
	ll val;
}edge[N << 1];
int head[N], tot;
int n;
int c[N];
ll dp[N], sz[N];
ll ans = inf;
void add(int u, int v, int l) {
	edge[++tot] = node{head[u], v, l};
	head[u] = tot;
}
void dfs1(int x, int fa) {
	sz[x] = c[x];
	for (int i = head[x]; i; i = edge[i].pre) {
		int y = edge[i].to;
		if (y == fa) continue;
		dfs1(y, x);
		sz[x] += sz[y];
		dp[x] += dp[y] + sz[y] * edge[i].val;
	}
}
void cut(int x, int y, int val) {
	dp[x] -= dp[y] + sz[y] * val;
	sz[x] -= sz[y];
}
void link(int x, int y, int val) {
	dp[x] += dp[y] + sz[y] * val;
	sz[x] += sz[y];
}
void change_root(int x, int y, int val) {
	cut(x, y, val);
	link(y, x, val);
}
void dfs2(int x, int fa) {
	cmin(ans, dp[x]);
	for (int i = head[x]; i; i = edge[i].pre) {
		int y = edge[i].to;
		if (y == fa) continue;
		change_root(x, y, edge[i].val);
		dfs2(y, x);
		change_root(y, x, edge[i].val);
	}
}
int main() {
	read(n);
	for (int i = 1; i <= n; i++) read(c[i]);
	for (int i = 1, a, b, v; i < n; i++) {
		read(a); read(b); read(v);
		add(a, b, v);
		add(b, a, v);
	}
	dfs1(1, 0);
	dfs2(1, 0);
	printf("%lld", ans);
	return 0;
}

[USACO10MAR]Great Cow Gathering G（换根dp）

标签：getc   oid   pac   long   problem   node   char   i++   print   

原文地址：https://www.cnblogs.com/zcr-blog/p/13205101.html