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HDU2120【并查集判环】
Ice_cream‘s world I
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 600 Accepted Submission(s): 344
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.
One answer one line.
Sample Input
8 100 11 21 32 43 40 55 66 73 64 7
Sample Output
3
大意：
题意很简单，刚开始我理解成两个塔之间有一个墙把他们隔开呢T_T
分析：并查集判环 只要在一个集合之中一定会有一块空地
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 6 const int maxn = 1005; 7 8 int fa[maxn]; 9 10 int find(int x) {11 if(fa[x] == x) return x;12 return fa[x] = find(fa[x]);13 }14 15 int main() {16 int n, m;17 int a, b;18 while(EOF != scanf("%d %d",&n, &m) ) {19 for(int i = 0; i < n; i++) fa[i] = i;20 int cnt = 0;21 while(m--) {22 scanf("%d %d",&a, &b);23 if(find(a) == find(b) ) {24 cnt++;25 } else {26 fa[find(a)] = find(b);27 }28 }29 printf("%d\n", cnt);30 }31 return 0;32 }
HDU2120【并查集判环】
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