杭电 2120 Ice_cream's world I
2015-07-29 20:10
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Ice_cream's world I
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 812 Accepted Submission(s): 476
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
在这道题中,通过判断根的相同与否来判断是否连成环。
#include<stdio.h> int pre[1010]; int find(int x){ int r,k,j; r=x; while(pre[r]!=r){ r=pre[r]; } k=x; while(k!=r){ j=pre[k]; pre[k]=r; k=j; } return r; } int main(){ int n,m; while(~scanf("%d%d",&n,&m)){ int i,x,y,c,d,count; for(i=0;i<n;i++) pre[i]=i; for(count=0,i=1;i<=m;i++){ scanf("%d%d",&x,&y); c=find(x); d=find(y); if(c!=d) { pre[d]=c; } else count++; } printf("%d\n",count); } return 0; }
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