zoukankan      html  css  js  c++  java
  • ACM MST Connect the Cities

    Connect the Cities
    Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

    Description

    In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.  
     

    Input

    The first line contains the number of test cases.
    Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
    To make it easy, the cities are signed from 1 to n.
    Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
    Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.
     

    Output

    For each case, output the least money you need to take, if it’s impossible, just output -1.
     

    Sample Input

    1 6 4 3 1 4 2 2 6 1 2 3 5 3 4 33 2 1 2 2 1 3 3 4 5 6
     

    Sample Output

    1
     
     
    好奇怪,g++交TLE了,c++叫过了
     
    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    #include <cmath>
    
    using namespace std;
    
    int t,n,m,k,sum;
    int father[505];
    struct Road
    {
        int a,b,value;
    } road[30000];
    int find(int x)
    {
        if(x!=father[x])
            father[x]=find(father[x]);
        return father[x];
    }
    void merge(int x,int y,int z,int &tp)
    {
        int a,b;
        a=find(x);
        b=find(y);
        if(a!=b)
        {
            tp+=z;
            father[a]=b;
        }
    }
    
    bool cmp(const Road &a,const Road &b)
    {
        return a.value<b.value;
    }
    
    
    
    int main()
    {
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d%d%d",&n,&m,&k);
            for(int i=1; i<=m; i++)
            {
                scanf("%d %d %d",&road[i].a,&road[i].b,&road[i].value);
            }
            int tt,tmp,x,y;
            int cnt = m;
            while(k--)
            {
                scanf("%d%d",&tt,&x);
                while(--tt)
                {
                    scanf("%d",&y);
                    road[++cnt].a = x;
                    road[cnt].b = y;
                    road[cnt].value = 0;;
                    x=y;
                }
            }
            sum = 0;
            sort(road+1,road+cnt+1,cmp);
    //        for(int i=1; i<=cnt; i++)
    //            printf("%d %d %d
    ",road[i].a,road[i].b,road[i].value);
            for(int i=1;i<=n;i++)
                father[i]=i;
            for (int i=1;i<=cnt;i++)
            {
                merge(road[i].a,road[i].b,road[i].value,sum);
            }
            int mother[505];
            for(int i=1; i<=n; i++)
            {
                mother[i] = find(i);
            }
            bool ok = 1;
            for(int i = 1; i<n; i++)
            {
                if(mother[i]!=mother[i+1])
                {
                    ok=0;
                    break;
                }
            }
            if(ok)
                printf("%d
    ",sum);
            else
                printf("-1
    ");
        }
    }
    

      

  • 相关阅读:
    系统并发报too much open files 错误
    plsql 安装Some Oracle Net versions cannot connect from a path with parentheses
    mysql登录报错
    web.xml配置文件详解之一servlet配置
    hibernate createQuery查询传递参数的两种方式
    mysql登录连接远程数据库命令行
    java 项目打包部署 过程
    flex ArrayCollection的新增与删除的同步
    加大eclipse以及jvm的内存
    2017 八月 UFED Series Releases 系列 6.3 重大更新发布
  • 原文地址:https://www.cnblogs.com/wejex/p/3261921.html
Copyright © 2011-2022 走看看