【LOJ】#2445. 「NOI2011」道路修建

题解

看完题目我的第一个反应是……要求最小花费的方案？！怎么求？？？

然后我把题读完了。好吧。

记录一下size就行，比NOIP普及组还要不如的题= =

代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#define enter putchar('
')
#define space putchar(' ')
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define eps 1e-7
#define MAXN 1000005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;char c = getchar();T f = 1;
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {putchar('-');x = -x;}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
int N;
struct node {
	int to,next,val;
}E[MAXN * 2];
int sumE,head[MAXN],siz[MAXN];
int64 ans;
void add(int u,int v,int c) {
	E[++sumE].to = v;
	E[sumE].next = head[u];
	E[sumE].val = c;
	head[u] = sumE;
}
int dfs(int u,int fa) {
	siz[u] = 1;
	for(int i = head[u] ; i ; i = E[i].next) {
		int v = E[i].to;
		if(v != fa) {
			dfs(v,u);
			siz[u] += siz[v];
			ans += 1LL * E[i].val * abs(N - siz[v] - siz[v]);
		}
	}
}
void Solve() {
	read(N);
	int u,v,c;
	for(int i = 1 ; i < N ; ++i) {
		read(u);read(v);read(c);
		add(u,v,c);add(v,u,c);
	}
	dfs(1,0);
	out(ans);enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}