HDU - 4424 Conquer a New Region(Kruskal的应用)

思路：类似Kruskal算法，将边按权值从大到小排序，每次加入一条边AB，如果以A所在的集合为中心城市，那么总权值为A所在集合的权值加上B所在集合点的数量乘以这条边的权值，因为A、B所在集合中的边肯定大于AB的权值，如果B所在的集合为中心城市，可以以同样的方法计算，取两者中大的那一个。

#include<cstdio>
#include<set>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<string>
#include<sstream>
#include<set>
#include<cmath>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 20;
const double EPS = 1e-5;
const int mod = 1e9 + 7;
typedef unsigned long long ull;
typedef long long LL;
int dx[] = {0, 0, -1, 1, -1, -1, 1, 1};
int dy[] = {1, -1, 0, 0, -1, 1, -1, 1};
struct Edge{
int from, to, c;
bool operator <(const Edge& rhs) const{
return c > rhs.c;
}
}a[maxn];
struct Node{
LL sum;
int num;
}b[maxn];
int f[maxn];
int Find(int x){
return x == f[x] ? x : f[x] = Find(f[x]);
}
int main(){
int n;
while(scanf("%d", &n) == 1){
for(int i = 0; i < n - 1; ++i){
scanf("%d%d%d", &a[i].from, &a[i].to, &a[i].c);
}
sort(a, a + n - 1);
for(int i = 1; i <= n; ++i) f[i] = i;
for(int i = 1; i <= n; ++i){
b[i].num = 1;
b[i].sum = 0;
}
for(int i = 0; i < n - 1; ++i){
int t1 = Find(a[i].from), t2 = Find(a[i].to);
if(b[t1].sum + (LL)b[t2].num * (LL)a[i].c > b[t2].sum + (LL)b[t1].num * (LL)a[i].c){
f[t2] = t1;
b[t1].num += b[t2].num;
b[t1].sum = b[t1].sum + (LL)b[t2].num * (LL)a[i].c;
}
else{
f[t1] = t2;
b[t2].num += b[t1].num;
b[t2].sum = b[t2].sum + (LL)b[t1].num * (LL)a[i].c;
}
}
printf("%lld\n", b[Find(1)].sum);
}
}
/*
3 3
3 1 2
3 2 0
2 3 2

*/