題解

看完題目我的第一個反應是……要求最小花費的方案？！怎麽求？？？

然後我把題讀完了。好吧。

記錄一下size就行，比NOIP普及組還要不如的題= =

代碼

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#define enter putchar(‘\n‘)
#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();
}

【LOJ】#2445. 「NOI2011」道路修建