hdu杭电2120 Ice_cream's world I【并查集】
2015-07-29 17:39
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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
Sample Output
//这题是找环的个数
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
//这题是找环的个数
#include<stdio.h> int per[100010]; int res=0; int find(int x) { if(x==per[x]) return x; return per[x]=find(per[x]); } void join(int x ,int y) { int fx=find(x); int fy=find(y); if(fx!=fy) { per[fx]=fy; } else { res++;//找出有几块地即找几个环 } } int main() { int a,b,m,n,i; while(~scanf("%d%d",&m,&n)) { for(i=0;i<m;++i) { per[i]=i; } res=0; while(n--) { scanf("%d%d",&a,&b); join(a,b); } printf("%d\n",res); } return 0; }
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