zoukankan      html  css  js  c++  java
  • Hdu 2120 Ice_cream's world I

    Ice_cream's world I

    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1611    Accepted Submission(s): 953

    Problem Description

    ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.

    Input

    In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between A and B has a wall(A and B are distinct). Terminate by end of file.

    Output

    Output the maximum number of ACMers who will be awarded.
    One answer one line.

    Sample Input

    8 10

    0 1

    1 2

    1 3

    2 4

    3 4

    0 5

    5 6

    6 7

    3 6

    4 7

    Sample Output

    3

    #include <stdio.h>
    int fa[1000+10];
    int n;
    void init()
    {
        for(int i=0;i<=n;i++)
        fa[i]=i;
    }
    int find(int x)
    {
        if(fa[x]!=x) fa[x]=find(fa[x]);
        return fa[x];
    }
    int main()
    {
        int m;
        while(~scanf("%d %d",&n,&m))
        {
            int result=0;
            init();
            for(int i=0;i<m;i++)
            {
                int a,b;
                scanf("%d %d",&a,&b);
                int x=find(a);
                int y=find(b);
                if(x!=y)
                {
                    fa[x]=y;
                }
                else//成环,但是不将这条边加到集合中去,记录成环一次             result++;
            }
            printf("%d
    ",result);
        }
        return 0;
    } 
    

      

  • 相关阅读:
    异步加载图片
    彩票项目
    linux 多线程的分离和可链接属性
    C库中system和atexit和exit的使用
    C库中getenv函数
    mode|平均数|方差|标准差|变异系数|四分位数|几何平均数|异众比率|偏态|峰态
    radar chart
    植物基因组|注释版本问题|重测序vs泛基因组
    signals function|KNN|SVM|average linkage|Complete linkage|single linkage
    supervised learning|unsupervised learning
  • 原文地址:https://www.cnblogs.com/zhangliu/p/7063202.html
Copyright © 2011-2022 走看看