POJ-2029 Get Many Persimmon Trees 树状数组

题意:给定一个矩形,现在给出这个矩形中某些点存在柿子树,问在一个长和宽限定的子矩形内最多有多少个柿子树.

解法:由于此题中数据量不大.直接树状数组统计然后暴力.

代码如下:

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int N, R, C, dx, dy;
int tr[105][105];

inline int lowbit(int x) {
    return x & -x;    
}

void modify(int x, int y, int val) {
    for (int i = x; i <= R; i+=lowbit(i)) {
        for (int j = y; j <= C; j+=lowbit(j)) {
            tr[i][j] += val;
        }
    }    
}

int sum(int x, int y) {
    int ret = 0;
    for (int i = x; i > 0; i-=lowbit(i)) {
        for (int j = y; j > 0; j-=lowbit(j)) {
            ret += tr[i][j];    
        }
    }
    return ret;
}

int get(int x, int y) {
    int ret, x0 = x-dx, y0 = y-dy;
    ret = sum(x, y) - sum(x0, y) - sum(x, y0) + sum(x0, y0);
    return ret;
}

int main() {
    int x, y, ret;
    while (scanf("%d", &N), N) {
        memset(tr, 0, sizeof (tr));
        ret = 0;
        scanf("%d %d", &C, &R); // 先读列再读行
        for (int i = 0; i < N; ++i) {
            scanf("%d %d", &y, &x);
            modify(x, y, 1);
        }
        scanf("%d %d", &dy, &dx);
        for (int i = dx; i <= R; ++i) {
            for (int j = dy; j <= C; ++j) {
                ret = max(ret, get(i, j));
            }    
        }
        printf("%d\n", ret);
    }
    return 0;
}