## ZOJ 3659 Conquer a New Region 第37届ACM/ICPC 长春赛区现场赛E题 （并查集）

Conquer a New Region

Time Limit: 5 Seconds      Memory Limit: 32768 KB

The wheel of the history rolling forward, our king conquered a new region in a distant continent.

There are N towns (numbered from 1 to N) in this region connected by several roads. It's confirmed that there is exact one route between any two towns. Traffic is important while controlled colonies are far away from the local country. We define the capacity C(i, j) of a road indicating it is allowed to transport at most C(i, j) goods between town i and town j if there is a road between them. And for a route between i and j, we define a value S(i, j) indicating the maximum traffic capacity between i and j which is equal to the minimum capacity of the roads on the route.

Our king wants to select a center town to restore his war-resources in which the total traffic capacities from the center to the other N - 1 towns is maximized. Now, you, the best programmer in the kingdom, should help our king to select this center.

#### Input

There are multiple test cases.

The first line of each case contains an integer N. (1 ≤ N ≤ 200,000)

The next N - 1 lines each contains three integers a, b, c indicating there is a road between town a and town b whose capacity is c. (1 ≤ a, b ≤ N, 1 ≤ c ≤ 100,000)

#### Output

For each test case, output an integer indicating the total traffic capacity of the chosen center town.

#### Sample Input

4
1 2 2
2 4 1
2 3 1
4
1 2 1
2 4 1
2 3 1


#### Sample Output

4
3


Contest: The 2012 ACM-ICPC Asia Changchun Regional Contest

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAXN=200010;

struct Node
{
int num;
long long sum;
}node[MAXN];

int F[MAXN];

int find(int x)
{
if(F[x]==-1)return x;
return F[x]=find(F[x]);
}

struct Edge
{
int a,b,c;
}edge[MAXN];

bool cmp(Edge a,Edge b)
{
return a.c>b.c;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<n-1;i++)
scanf("%d%d%d",&edge[i].a,&edge[i].b,&edge[i].c);
sort(edge,edge+n-1,cmp);

for(int i=1;i<=n;i++)
{
node[i].num=1;
node[i].sum=0;
F[i]=-1;
}

for(int i=0;i<n-1;i++)
{
int a=edge[i].a;
int b=edge[i].b;
int t1=find(a);
int t2=find(b);
if(node[t1].sum+(long long)edge[i].c*node[t2].num<node[t2].sum+(long long)edge[i].c*node[t1].num)
{
F[t1]=t2;
node[t2].num+=node[t1].num;
node[t2].sum+=(long long)edge[i].c*node[t1].num;
}
else
{
F[t2]=t1;
node[t1].num+=node[t2].num;
node[t1].sum+=(long long)edge[i].c*node[t2].num;
}
}
printf("%lld\n",node[find(1)].sum);

}
return 0;
}

posted on 2012-10-17 08:54  kuangbin  阅读(1296)  评论(0编辑  收藏  举报

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