hdoj 2120 Ice_cream's world I
2015-07-29 19:03
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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): 814 Accepted Submission(s): 478
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
Output the maximum number of ACMers who will be awarded.
One answer one line.
[align=left]Sample Input[/align]
8 10 0 1 1 2 1 3 2 4 3 4 0 5 5 6 6 7 3 6 4 7
[align=left]Sample Output[/align]
3
思路:求环的个数。输入数据有多组。
代码:
#include<stdio.h> #include<stdlib.h> #include<algorithm> #include<queue> using namespace std; int per[1100]; int n,m,f; int init() { for(int i=0;i<n;i++) { per[i]=i; } } int find(int x) { int r; r=x; while(r!=per[r]) { r=per[r]; } per[x]=r; return r; } int join(int x,int y) { int fx=find(x); int fy=find(y); if(fx!=fy) per[fx]=fy; else f++; } int main() { int a,b,i; while(scanf("%d%d",&n,&m)!=EOF) { f=0; init(); for(i=0;i<m;i++) { scanf("%d%d",&a,&b); join(a,b); } printf("%d\n",f); } return 0; }
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