[luoguP1131] [ZJOI2007]时态同步（贪心）

传送门

显然是一棵树。

又显然一段一段地增加比较优。

我们可以dfs，并且尽量把每一个节点所有子树中所有节点的时间增加到一样。

#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#define N 500001
#define LL long long

using namespace std;

LL ans;
int n, s, cnt;
int head[N], to[N << 1], val[N << 1], nex[N << 1], dis[N];
bool vis[N];
vector <int> q[N];

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

inline void add(int x, int y, int z)
{
	to[cnt] = y;
	val[cnt] = z;
	nex[cnt] = head[x];
	head[x] = cnt++;
}

inline void dfs(int u)
{
	int i, v, t = -1;
	vis[u] = 1;
	for(i = head[u]; ~i; i = nex[i])
	{
		v = to[i];
		if(!vis[v])
		{
			dis[v] = dis[u] + val[i];
			dfs(v);
			t = max(t, dis[v]);
			q[u].push_back(dis[v]);
		}
	}
	if(t > 0)
	{
		dis[u] = t;
		for(i = 0; i < q[u].size(); i++) ans += t - q[u][i];
	}
}

int main()
{
	int i, x, y, z;
	n = read();
	s = read();
	memset(head, -1, sizeof(head));
	for(i = 1; i < n; i++)
	{
		x = read();
		y = read();
		z = read();
		add(x, y, z);
		add(y, x, z);
	}
	dfs(s);
	printf("%lld\n", ans);
	return 0;
}