ZJNU 1333 - senior second question blocks--

Because we need to put a judge once every run again full picture bfs is clearly unrealistic

And because only three points, no black and white

So you can use disjoint-set optimization

Add a piece it is judged that the situation around the four groups of

Discrimination between situations arise and answer relationship Note

 1 /*
 2 Written By StelaYuri
 3 */
 4 #include<stdio.h>
 5 #include<memory.h>
 6 int N,M,dx[4]={1,0,-1,0},dy[4]={0,1,0,-1},gp[250010];
 7 char cm[505][505];
 8 int findp(int p){
 9     return p==gp[p]?p:(gp[p]=findp(gp[p]));
10 }
11 int prime(int x,int y){
12     return x>=0&&y>=0&&x<N&&y<N;
13 }
14 int main(){
15     int i,j,X,Y,xx,yy,ans=0,g,d1,d2;
16     char C;
17     memset(cm,'.',sizeof cm);
18     for(i=0;i<250000;i++)
19         gp[i]=i;
20     scanf("%d%d",&N,&M);
21     for(i=0;i<M;i++){
22         scanf("%*c%c%d%d",&C,&X,&Y);
23         ans++;
24         for(g=j=0;j<4;j++){
25             xx=X-1+dx[j];
26             yy=Y-1+dy[j];
27             if(prime(xx,yy)&&cm[xx][yy]==C){
28                 findp = D1 (GP [(X- . 1 ) * N + (Y- . 1 )]);
 29                  D2 = findp (GP [XX * N + YY]);
 30                  IF (! G) { // four surrounding positions If further is not found, may be combined one with the set of direct and reducing a set of answers 
31 is                      G = . 1 ;
 32                      IF (! D1 = D2) {
 33 is                          IF (D1 < D2)
 34 is                              GP [D2] = D1;
 35                          the else 
36                              GP [D1 ] = D2;
 37 [                      }
 38 is                      ans-- ;
 39                 }
 40                  the else { // four surrounding positions if a same group has been found, and only when the different sets of points where the two were combined and reduced need for a answer group 
41 is                      IF (D1! = D2) {
 42 is                          IF (D1 < D2)
 43 is                              GP [D2] = D1;
 44 is                          the else 
45                              GP [D1] = D2;
 46 is                          ans-- ;
 47                      }
 48                  }
 49              }
 50          }
 51 is          cm & lt [X- . 1 ] [Y- . 1 ] = C;
 52 is         printf("%d\n",ans);
53     }
54     
55     return 0;
56 }