Ice_cream‘s world I
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 698 Accepted Submission(s): 398
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
题意:就是图中有几个环。
#include<stdio.h> const int N = 1015; int fath[N],n; void init() { for(int i =0;i<=n;i++) fath[i]=i; } int findfath(int x) { if(x==fath[x]) return fath[x]; fath[x]=findfath(fath[x]); return fath[x]; } int setfath(int x,int y) { x=findfath(x); y=findfath(y); if(x==y) return 1; fath[x]= y; return 0; } int main() { int x,y,m,ans; while(scanf("%d%d",&n,&m)>0) { init(); ans=0; while(m--) { scanf("%d%d",&x,&y); ans+=setfath(x,y); } printf("%d\n",ans); } }
HDU2120Ice_cream's world I(基础并查集)