ZOJ Conquer a New Region（并查集）

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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

解题思路：

贪心 + 并查集， 从大到小处理边，保证当前的边是已经处理过的边中权值最小的一个，然后判断是加到起点还是终点。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#define LL long long
using namespace std;
const int MAXN = 200000 + 10;
int F[MAXN];
struct Edge
{
    int u, v;
    int c;
    bool operator < (const Edge& rhs)const
    {
        return c > rhs.c;
    }
} edge[MAXN];
struct Node
{
    int num;
    long long sum;
} P[MAXN];
int Find(int x)
{
    return F[x] == -1 ? x : F[x] = Find(F[x]);
}
int main()
{
    int N;
    while(scanf("%d", &N)!=EOF)
    {
        //int u, v, c;
        for(int i=1; i<N; i++)
            scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].c);
        for(int i=1; i<=N; i++)
        {
            F[i] = -1;
            P[i].num = 1;
            P[i].sum = 0;
        }
        sort(edge + 1, edge + N);
        //for(int i=1;i<N;i++)
        //  cout << edge[i].u << ' ' << edge[i].v << ' ' <<edge[i].c << endl;
        for(int i=1; i<N; i++)
        {
            int fx = Find(edge[i].u);
            int fy = Find(edge[i].v);
            if(P[fx].sum + (long long )P[fy].num * edge[i].c < P[fy].sum + (long long )P[fx].num * edge[i].c)
            {
                F[fx] = fy;
                P[fy].num += P[fx].num;
                P[fy].sum += (long long)P[fx].num * edge[i].c;
            }
            else
            {
                F[fy] = fx;
                P[fx].num += P[fy].num;
                P[fx].sum += (long long )P[fy].num * edge[i].c;
            }
        }
        long long ans = P[Find(1)].sum;
        printf("%lld\n", ans);
    }
    return 0;
}

ZOJ Conquer a New Region（并查集）

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原文地址：http://blog.csdn.net/moguxiaozhe/article/details/45103669