Hdoj Ice_cream ' s World I 2120 "and check the set judgment into a ring"

Ice_cream ' s World I Time limit:3000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 848 Accepted Submission (s): 494


Problem Descriptionice_cream ' s world is a rich country, it had many fertile lands. Today, the Queen of Ice_cream wants award land to diligent acmers. So there is some watchtowers is set up, and wall between watchtowers is build, in order to partition the Ice_cream ' s WOR Ld. But what many acmers at most can be awarded by the Queen is a big problem. One wall-surrounded land must is given to only one acmer and no walls is crossed if you can help the Queen solve this PR Oblem, you'll be get-a land.
Inputin the case, first-integers N, M (n<=1000, m<=10000) is represent the number of watchtower and the number O F Wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain, integers a, b mean between A and b have A wall (A and B are distinct). Terminate by end of file.
Outputoutput the maximum number of acmers who'll be awarded.
One answer one line.
Sample Input

8 100 11 21 32 43 40 55 66 73 64 7

Sample Output

3

Authorwiskey
Sourcehdu 2007-10 Programming Contest_warmup
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#include <iostream> #include <stdio.h> #include <algorithm>using namespace std;int r[1010];int flag; int find_x (int x) {while (x!=r[x]) X=r[x];return x;} void Fun (int x,int y) {x=find_x (x); y=find_x (y); if (x!=y) {r[y]=x;} else flag++;} int main () {    int n,m;    while (scanf ("%d%d", &n,&m)!=eof) {for (int i=0;i<=n;i++) r[i]=i;flag=0;int a,b;for (int i=0;i<m;i++) { scanf ("%d%d", &a,&b); fun (b);} printf ("%d\n", flag);    }    return 0;}

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Hdoj Ice_cream ' s World I 2120 "and check the set to determine the ring"