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; }