hdu2120 Ice_cream's world I (并查集查找环的个数)
2015-08-13 18:28
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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): 887 Accepted Submission(s): 517
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
Author
Wiskey
Source
HDU 2007-10 Programming Contest_WarmUp
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Statistic | Submit | Discuss | Note
这道题的意思就是查找一个图里面有几个环。
怪我一切知识都学的太死了。唉,起初竟然没想到。。多希望再来几个这样的题
让我捡漏。。
#include <stdio.h> int fa[1005]; int find(int x) { if(fa[x]!=x) fa[x]=find(fa[x]); return fa[x]; } bool comb(int a,int b) { a=find(a); b=find(b); if(a==b)//如果根相等就是出现了环 return true; else { fa[a]=b; return false; } } void init(int n) { for(int i=0;i<n;i++) fa[i]=i; } int main() { int n,k; while(scanf("%d %d",&n,&k)!=EOF) { init(n); int sum=0; for(int i=0;i<k;i++) { int a,b; scanf("%d %d",&a,&b); if(comb(a,b)) sum++; } printf("%d\n",sum); } return 0; }
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