zoukankan      html  css  js  c++  java
  • Civil and Evil Engineer(普林姆)

    http://acm.sdut.edu.cn:8080/vjudge/contest/view.action?cid=198#problem/E

    水题一道,题意就是让求一遍最小生成树与最大生成树,但我因为Each case contains a blank line and an integer n (1 ≤ n ≤ 100) denoting the number of houses.这句话错了三遍,有两种解决方案,一不用管空格,二gets(str);

    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    #include <stdlib.h>
    #include <math.h>
    #define N 1000001
    int map[101][101],map1[101][101];
    int v[101],v1[101];
    int dis[101],dis1[101];
    int n;
    int Primmin()
    {
        int min,k,sum=0;
        memset(v,0,sizeof(v));
        v[0]=1;
        for(int i=0;i<=n;i++)
          dis[i]=map[0][i];
        for(int i=0;i<n;i++)
        {
            min=N;
            for(int j=0;j<=n;j++)
            {
                if(v[j]==0&&min>dis[j])
                {
                    min=dis[j];
                    k=j;
                }
            }
            v[k]=1;
            sum=sum+min;
            for(int j=0;j<=n;j++)
            {
                if(v[j]==0&&dis[j]>map[k][j])
                {
                    dis[j]=map[k][j];
                }
            }
        }
        return sum;
    }
    int Primmax()
    {
        int min,k,sum=0;
        memset(v1,0,sizeof(v1));
        v1[0]=1;
        for(int i=0;i<=n;i++)
          dis1[i]=map1[0][i];
        for(int i=0;i<n;i++)
        {
            min=0;
            for(int j=0;j<=n;j++)
            {
                if(v1[j]==0&&min<dis1[j])
                {
                    min=dis1[j];
                    k=j;
                }
            }
            sum=sum+min;
            v1[k]=1;
            for(int j=0;j<=n;j++)
            {
                if(v1[j]==0&&dis1[j]<map1[k][j])
                {
                    dis1[j]=map1[k][j];
                }
            }
        }
        return sum;
    }
    int main()
    {
        int T,K=0;
        char ch[101];
        scanf("%d",&T);
        while(T--)
        {
            K++;
            gets(ch);
            scanf("%d",&n);
            for(int i=0;i<=n;i++)
            {
                for(int j=0;j<=n;j++)
                {
                    map[i][j]=N;
                    map[j][i]=N;
                }
                map[i][i]=0;
            }
            for(int i=0;i<=n;i++)
            {
                for(int j=0;j<=n;j++)
                {
                    map1[i][j]=0;
                    map1[j][i]=0;
                }
            }
            int xx,yy,zz;
            while(scanf("%d%d%d",&xx,&yy,&zz)!=EOF)
            {
                if(xx==0&&yy==0&&zz==0)  break;
                if(map[xx][yy]>zz)
                {
                    map[xx][yy]=zz;
                    map[yy][xx]=zz;
                }
                if(map1[xx][yy]<zz)
                {
                    map1[xx][yy]=zz;
                    map1[yy][xx]=zz;
                }
            }
            int sum1=Primmin();
            int sum2=Primmax();
             long long int tt=sum1+sum2;
            if(tt%2==0)
            {
                 tt=tt/2;
                 printf("Case %d: %lld
    ",K,tt);
            }
            else printf("Case %d: %lld/2
    ",K,tt);
        }
        return 0;
    }
  • 相关阅读:
    CF1096C Polygon for the Angle
    CF1149A Prefix Sum Primes
    CF1153C Serval and Parenthesis Sequence
    leetcode134 Gas Station
    leetcode23 Merge k Sorted Lists
    leetcode982 Triples with Bitwise AND Equal To Zero
    CF1110C Meaningless Operations
    【bzoj4567】SCOI2016背单词
    【不定期更】【友链整理】
    【bzoj2006】NOI2010超级钢琴
  • 原文地址:https://www.cnblogs.com/zhangmingcheng/p/3887062.html
Copyright © 2011-2022 走看看