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  • poj 3723 Conscription

    Description

    Windy has a country, and he wants to build an army to protect his country. He has picked up N girls and M boys and wants to collect them to be his soldiers. To collect a soldier without any privilege, he must pay 10000 RMB. There are some relationships between girls and boys and Windy can use these relationships to reduce his cost. If girl x and boy y have a relationship d and one of them has been collected, Windy can collect the other one with 10000-d RMB. Now given all the relationships between girls and boys, your assignment is to find the least amount of money Windy has to pay. Notice that only one relationship can be used when collecting one soldier.

    Input

    The first line of input is the number of test case.
    The first line of each test case contains three integers, N, M and R.
    Then R lines followed, each contains three integers xi, yi and di.
    There is a blank line before each test case.

    1 ≤ N, M ≤ 10000
    0 ≤ R ≤ 50,000
    0 ≤ xi < N
    0 ≤ yi < M
    0 < di < 10000

    Output

    For each test case output the answer in a single line.

    Sample Input

    2
    
    5 5 8
    4 3 6831
    1 3 4583
    0 0 6592
    0 1 3063
    3 3 4975
    1 3 2049
    4 2 2104
    2 2 781
    
    5 5 10
    2 4 9820
    3 2 6236
    3 1 8864
    2 4 8326
    2 0 5156
    2 0 1463
    4 1 2439
    0 4 4373
    3 4 8889
    2 4 3133
    

    Sample Output

    71071
    54223
    

    Source

     
    最小生成树,Kruskall算法。
    代码:
    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <algorithm>
    using namespace std;
    struct rela {
        int x,y,d;
    }s[50000];
    int t,n,m,r;
    int f[20000];
    void init() {
        for(int i = 0;i < (n + m);i ++) {
            f[i] = i;
        }
    }
    int getf(int x) {
        if(x != f[x])f[x] = getf(f[x]);
        return f[x];
    }
    int mer(int x,int y) {
        int xx = getf(x);
        int yy = getf(y);
        if(xx == yy)return 1;
        f[xx] = yy;
        return 0;
    }
    bool cmp(rela a,rela b) {
        return a.d > b.d;
    }
    int main() {
        scanf("%d",&t);
        while(t --) {
            scanf("%d%d%d",&n,&m,&r);
            int ans = (n + m) * 10000;
            init();
            for(int i = 0;i < r;i ++) {
                scanf("%d%d%d",&s[i].x,&s[i].y,&s[i].d);
                s[i].y += n;
            }
            sort(s,s + r,cmp);
            for(int i = 0;i < r;i ++) {
                if(mer(s[i].x,s[i].y))continue;
                ans -= s[i].d;
            }
            printf("%d
    ",ans);
        }
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/8023spz/p/9170423.html
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