并查集hdu4424

Conquer a New Region

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1271    Accepted Submission(s): 415

Problem Description

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
题意：给出一个树形图和边权值，边权值是容量，问把某个点作为首都，向其他n-1个点运输货物，货物容量不超过路径上的边权值；问可以输出的最大货物是多少？
分析：并查集，按照边权从大到小排序，在加入该边的同时，比较把两个集合合并到那个集合更优：

#include"string.h"
#include"stdio.h"
#include"iostream"
#include"algorithm"
#include"queue"
#include"stack"
#include"stdlib.h"
#include"map"
#include"string"
#include"math.h"
#define inf 10000000
#define INF 0x3f3f3f3f
const double PI=acos(-1.0);
const double r2=sqrt(2.0);
const int M=200009;
const int N=1010*502*2;
const double g=9.8;
#define eps 1e-10
using namespace std;
int f[M],h[M];
__int64 sum[M],ans;
int finde(int x)
{
    if(x!=f[x])
        f[x]=finde(f[x]);
    return f[x];
}
void make(int a,int b,__int64 w)
{
    int x=finde(a);
    int y=finde(b);
    __int64 f1,f2;
    f1=sum[x]+h[y]*w;
    f2=sum[y]+h[x]*w;
    if(f1>f2)
    {
        f[y]=x;
        sum[x]+=h[y]*w;
        h[x]+=h[y];
    }
    else
    {
        f[x]=y;
        sum[y]+=h[x]*w;
        h[y]+=h[x];
    }
}
struct node
{
    int u,v;
    __int64 w;
}e[M];
int cmp(node a,node b)
{
    return a.w>b.w;
}
int main()
{
    int n,i;
    while(scanf("%d",&n)!=-1)
    {
        for(i=1;i<=n;i++)
        {
            f[i]=i;
            sum[i]=0;
            h[i]=1;
        }
        ans=0;
        for(i=1;i<n;i++)
            scanf("%d%d%I64d",&e[i].u,&e[i].v,&e[i].w);
        sort(e+1,e+n,cmp);
        for(i=1;i<n;i++)
            make(e[i].u,e[i].v,e[i].w);
        printf("%I64d\n",sum[finde(1)]);
    }
    return 0;
}