【题解】LuoGu1131：[ZJOI2007]时态同步

原题传送门
这道题目可以放到普及T4难度
树型DP，记两个数组
$dp[u]$表示处理完自己为根的子树所需代价
$len[u]$表示处理好自己的儿子后的最大深度
转移方程：$dp[u]={\sum }_{}^{}dp[v]+{\sum }_{}^{}(max(len[v]+w[u][v])-(len[v]+w[u][v])),(v=u_{son})$

坑点是要开longlong，我因此wa了两发

Code：

#include <bits/stdc++.h>
#define maxn 1000010
#define LL long long
using namespace std;
struct Edge{
	int to, next, len;
}edge[maxn << 1];
int num, head[maxn], n, r;
LL dp[maxn], len[maxn];

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48) ;
	return s * w;
}

void addedge(int x, int y, int z){ edge[++num] = (Edge) { y, head[x], z }; head[x] = num; }

void dfs(int u, int pre){
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if (v != pre){
			dfs(v, u);
			dp[u] += dp[v], len[u] = max(len[u], len[v] + edge[i].len);
		}
	}
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if (v != pre) dp[u] += len[u] - len[v] - edge[i].len;
	}
}

int main(){
	n = read(), r = read();
	for (int i = 1; i < n; ++i){
		int x = read(), y = read(), z = read();
		addedge(x, y, z); addedge(y, x, z);
	}
	dfs(r, 0);
	printf("%lld\n", dp[r]);
	return 0;
}