Check the HDU 2120

Tag: and query set

In fact, it is to determine whether the newly added vertex is in the same set.

Code abuse me a thousand times, I am waiting for the first love of code

Ice_cream's world I Time Limit: 3000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 583 accepted submission (s): 333


Problem descriptionice_cream's world is a rich country, it has alias 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 Wils are crossed, if you can help the Queen solve this problem, you will be get a land.
Inputin 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.
Outputoutput the maximum number of acmers who will be awarded.
One answer one line.
Sample Input

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

Sample output

 
3
 

# Include <stdio. h> # include <stdlib. h> # include <malloc. h> # include <limits. h> # include <ctype. h> # include <string. h> # include <string> # include <queue> # include <algorithm> # include <iostream> # include <stack> # include <deque> # include <vector> # include <set> # include <map> using namespace STD; # define maxn 1000 + 10 # define maxn1 10000 + 10int father [maxn]; int A [maxn1]; int B [maxn1]; int find (INT X) {If (X! = Father [x]) {FATHER [x] = find (father [x]);} return father [X];} int main () {int n, m; int I; while (~ Scanf ("% d", & N, & M) {for (I = 0; I <maxn; I ++) {FATHER [I] = I ;} memset (A, 0, sizeof (a); memset (B, 0, sizeof (B); for (I = 0; I <m; I ++) {scanf ("% d", & A [I], & B [I]) ;}int sum = 0; for (I = 0; I <m; I ++) {int F1 = find (A [I]); int F2 = find (B [I]); If (F1! = F2) {If (F1 <F2) {FATHER [F2] = F1;} else {FATHER [F1] = F2 ;}} else {sum ++ ;}} printf ("% d \ n", sum);} return 0 ;}

Check the HDU 2120