HDU-2120-Ice_cream's world I【并查集】
2016-08-01 20:12
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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): 1265 Accepted Submission(s): 758
[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
题解:求由 0 到 n-1 围成的大区域被墙给分成了几个小区域,上图:
#include<cstdio> #include<algorithm> const int Q=1e4; int n,m,fa[Q+10]; int find(int x) { if(x==fa[x]) return x; return fa[x]=find(fa[x]); } void Union(int x,int y) { int nx=find(x); int ny=find(y); if(nx!=ny) { fa[ny]=nx; } } int main() { while(~scanf("%d %d",&n,&m)) { for(int i=0;i<n;i++) { fa[i]=i; } int a,b,ans=0; while(m--) { scanf("%d %d",&a,&b); if(find(a)==find(b)) { ans++; } Union(a,b); } printf("%d\n",ans); } return 0; }
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