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  • HDU 2120 Ice_cream's world I

     

                  Ice_cream's world I



     

    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
     
     1 #include<cstdio>
     2 using namespace std;
     3 
     4 int bin[1005];
     5 int count;
     6 
     7 int find(int x)
     8 {
     9     int r=x;
    10     while(bin[r]!=r)
    11     r=bin[r];
    12     int temp=x;
    13     while(temp!=r)
    14     {
    15         int j=bin[temp];
    16         bin[temp]=r;
    17         temp=j;
    18     }
    19     return r;
    20 }
    21 
    22 void merge(int x,int y)
    23 {
    24     int fx=find(x);
    25     int fy=find(y);
    26     if(fx!=fy)
    27     bin[fy]=fx;
    28     else
    29     count++;
    30 }
    31 
    32 int main()
    33 {
    34     //freopen("in.txt","r",stdin);
    35     int m,n,a,b;
    36     while(scanf("%d%d",&m,&n)!=EOF)
    37     {
    38         count=0;
    39         for(int i=0;i<m;i++)
    40         bin[i]=i;
    41         for(int i=0;i<n;i++)
    42         {
    43             scanf("%d%d",&a,&b);
    44             merge(a,b);
    45         }
    46         printf("%d
    ",count);
    47     }
    48     return 0;
    49 }
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  • 原文地址:https://www.cnblogs.com/homura/p/4686931.html
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