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2015
07-16

HDU 4424-Conquer a New Region-栈-[解题报告]HOJ

Conquer a New Region

问题描述 :

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.

输入:

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)

输出:

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)

样例输入:

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

样例输出:

4
3

/*
分析:
    并查集(2012长春现场赛E题)。
    蛋疼的再次爆栈了,囧~~
    意外的在Statistic排了第一耶,625MS。太意外了,囧~~~
    很巧妙的想法,由于路径的容量只与最小边有关,所
以把边从大到小排序,然后并查集,合并两个集合的时候
取max,就行了。
   
                                              2012-10-24
*/

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include"stdio.h"
#include"string.h"
#include"stdlib.h"
struct Eage
{
    int a,b;
    __int64 len;
}eage[200111];
int pre[200111];
int num[200111];
__int64 sum[200111];
int cmp(const void *a,const void *b)
{
    struct Eage *c,*d;
    c=(struct Eage *)a;
    d=(struct Eage *)b;
    return d->len-c->len;
}
void build(int n)
{
    int i;
    for(i=1;i<=n;i++)   {pre[i]=i;num[i]=1;sum[i]=0;}
}
int find(int k)
{
    if(pre[k]==k)   return k;
    pre[k]=find(pre[k]);
    return pre[k];
}
void Union(int f1,int f2,__int64 dir)
{
    pre[f1]=f2;
    num[f2]+=num[f1];
    sum[f2]+=dir;
}
int main()
{
    int n,t;
    int i,l;
    int f1,f2;
    long long dir1,dir2;
    while(scanf("%d",&n)!=-1)
    {
        build(n);
        t=n-1;
        for(i=0;i<t;i++)    scanf("%d%d%I64d",&eage[i].a,&eage[i].b,&eage[i].len);
        qsort(eage,t,sizeof(eage[0]),cmp);

        for(i=0;i<t;i++)
        {
            f1=find(eage[i].a);
            f2=find(eage[i].b);
            if(f1==f2)  continue;
			dir1=num[f2]*eage[i].len+sum[f1];
			dir2=num[f1]*eage[i].len+sum[f2];
            if(dir1>dir2) Union(f2,f1,dir1-sum[f1]);
            else          Union(f1,f2,dir2-sum[f2]);
        }

        printf("%I64d\n",sum[find(1)]);
    }
    return 0;
}

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参考:http://blog.csdn.net/ice_crazy/article/details/8105530