zoukankan      html  css  js  c++  java
  • CDOJ 145 Earthquake

    After the big earthquake, a lot of roads have been destroyed, some towns are disconnected with each other.

    In order to save the trapped as soon as possible, we need to try our best to rebuild the roads, and make sure all the towns will be reconnected(that is any villages is connected to the others with a clear route at least).

    Unfortunately, we have only one team to rebuild the roads. Now,please tell us how long do you think these roads can be reconnected.

    Input

    The first line contains a number T denotes the number of test case.

    For each test case,

    In the first line, you will get two number N (1N1000) and M(1MN×N), denotes the number of towns and the number of roads.

    The next M lines, each contains three number A,B,C, denotes there is a road between A and B that needed C (1×C×1000) minutes to rebuild.

    Output

    For each test case, you should output a line contains a number denotes the minimal time need to rebuild the roads so that all the towns are connected.

    Sample input and output

    Sample InputSample Output
    2
    3 3
    1 2 3
    2 3 3
    3 1 7
    3 3
    1 2 3
    2 3 3
    3 1 1
    6
    4
    #include <iostream>
    #include <algorithm>
    using namespace std;
    
    int V,E,v[1005];
    class edge {
    public:
        int weight,from,to;
    }e[1000005];
    
    bool cmp(edge a,edge b) {
        return a.weight<b.weight;
    }
    
    int find(int x) {
        return (v[x]==x)?x:(v[x]=find(v[x]));
    }
    
    void MST() {
        int vnum=0,all_weight=0,x,y;
        for (int i=0; i<E&&vnum<V-1; i++) {
            x=find(e[i].from);
            y=find(e[i].to);
            if (x!=y) {
                v[x]=y;
                all_weight+=e[i].weight;
                vnum++;
            }
        }
        printf("%d
    ",all_weight);
    }
    
    int main() {
        int t;
        scanf("%d",&t);
        while (t--) {
            scanf("%d%d",&V,&E);
            for (int i=0; i<V; i++)
                v[i]=i;
            for (int i=0; i<E; i++) {
                scanf("%d%d%d",&e[i].from,&e[i].to,&e[i].weight);
                e[i].from--;
                e[i].to--;
            }
            sort(e, e+E, cmp);
            MST();
        }
    }
  • 相关阅读:
    jquery跨域3
    juery的跨域请求2
    jquery的跨域请求
    synchronized与Lock的区别
    springboot之启动原理解析及源码阅读
    java中Number类理解
    springboot中配置文件application.properties的理解
    restTemplate设置访问超时
    BigDecimal.setScale 处理java小数点
    NIO之FileChannel类的理解和使用
  • 原文地址:https://www.cnblogs.com/soildom/p/7979192.html
Copyright © 2011-2022 走看看