poj 2029 Get Many Persimmon Trees

来源：互联网发布：mac版网游加速器编辑：程序博客网时间：2024/05/20 07:18

本题可以二维树状数组的思想来解决，暴力好像也可以，数据比较小。用二维树状数组，就是枚举指定长宽的矩阵内*号的个数，并求出数目最大的。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 110;int c[MAXN][MAXN];int lowbit(int x){     return x & (-x);}void Modefy(int x, int y, int data){     for(int i = x; i < MAXN; i += lowbit(i))          for(int j = y; j < MAXN; j += lowbit(j))               c[i][j] += data;}int GetSum(int x, int y){     int sum = 0;     for(int i = x; i >0; i -= lowbit(i))          for(int j = y; j > 0; j -= lowbit(j))               sum += c[i][j];     return sum;}int Querry(int x1, int y1, int x2, int y2){     return GetSum(x2, y2) - GetSum(x1-1,y2) - GetSum(x2, y1-1) + GetSum(x1-1,y1-1);}int main(){     int n;     int x, y;     int w, h;     int S, T;     while(scanf("%d", &n) != EOF && n)     {          memset(c, 0, sizeof(c));          scanf("%d %d", &w, &h);          for(int i = 1; i <= n; ++i)          {               scanf("%d %d", &x, &y);               Modefy(x, y, 1);          }          scanf("%d %d", &S, &T);          int ans = 0;          for(int i = S; i <= w; ++i)               for(int j = T; j <= h; ++j)               {                    int sum = Querry(i-S+1, j-T+1, i, j);                    if(sum > ans)                         ans = sum;               }          printf("%d\n", ans);     }     return 0;}