POJ 2029 Get Many Persimmon Trees（DP）

 1 //2029
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <iostream>
 5 
 6 using namespace std ;
 7 
 8 int mapp[110][110] ;
 9 
10 int main()
11 {
12     int N ,S,T ,W,H;
13     while(scanf("%d",&N) != EOF && N)
14     {
15         int x,y ;
16         memset(mapp,0,sizeof(mapp)) ;
17         scanf("%d %d",&W,&H) ;
18         for(int i = 0 ; i < N ; i++)
19         {
20             scanf("%d %d",&x,&y) ;
21             mapp[x][y] = 1 ;
22         }
23         scanf("%d %d",&S,&T) ;
24         for(int i = 1 ; i <= W ; i++)
25             for(int j = 1 ; j <= H ; j++)
26                 mapp[i][j] += (mapp[i-1][j]+mapp[i][j-1]-mapp[i-1][j-1]) ;
27         int ans = -1 ;
28         for(int i = S ; i <= W ;i ++)
29         {
30             for(int j = T ; j <= H ; j++)
31             {
32                 ans = max(ans,mapp[i][j]-mapp[i-S][j]-mapp[i][j-T]+mapp[i-S][j-T]) ;
33             }
34         }
35         printf("%d\n",ans) ;
36     }
37     return 0 ;
38 }

 1 #include<iostream>
 2 #include<math.h>
 3 #include<stdio.h>
 4 #include<string.h>
 5 #define MAX 102
 6 using namespace std;
 7 int c[MAX][MAX];
 8 int lowbit(int x)
 9 {
10     return x & (-x);
11 }
12 void add( int x,int y)
13 {
14     int i=x,j=y;
15       for(i=x;i<MAX;i+=lowbit(i))
16           for(j=y;j<MAX;j+=lowbit(j))
17               c[i][j]++;
18 }
19 int sum(int x,int y)
20 {
21    int i,k,sum = 0;
22     for(i=x; i>0; i-=lowbit(i))
23         for(k=y; k>0; k-=lowbit(k))
24             sum += c[i][k];
25     return sum;
26 }
27 int main()
28 {
29     int w,h,i,j,s,t,n;
30     int x,y,ans;
31     while(scanf("%d",&n))
32     {
33         if(n==0)break;
34         memset(c,0,sizeof(c));
35         scanf("%d%d",&w,&h);
36         for(i=0;i<n;i++)
37         {
38             scanf("%d%d",&x,&y);
39             add(x,y);
40         }
41         scanf("%d%d",&s,&t);
42         ans=0;
43         for(i=s;i<=w;i++)
44         {
45             for(j=t;j<=h;j++)
46             {
47                 ans=max(ans,sum(i,j)-sum(i-s,j)-sum(i,j-t)+sum(i-s,j-t));
48             }
49         }
50         printf("%d\n",ans);
51     }
52 }