枚举_百炼 2812 讨厌的青蛙

#define _CRT_SECURE_NO_WARNINGS  
#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <stdlib.h>
#include <vector>
#include <map>
#include <queue>
#include <string>
#include <iostream>
#include <ctype.h>
#include <string.h>
#include <set>
#include <stack>
#include<functional>
using namespace std;
#define Size 5005
#define maxn  1<<30
#define minn  1e-6
/*
枚举的剪枝
*/
struct E{
    int x, y;
}a[Size];
int row, col,n;
int  cmp(const void*t, const void *s){
    E *p1, *p2;
    p1 = (E*)t;
    p2 = (E*)s;
    if (p1->x != p2->x) return p1->x - p2->x;
    else return p1->y - p2->y;
}
int search(E t, int dx, int dy){
    E temp;
    temp.x = t.x + dx;
    temp.y = t.y + dy; 
    int steps = 2;
    while (temp.x > 0 && temp.x <= row && temp.y > 0 && temp.y <= col){
        if (!bsearch(&temp,a,n,sizeof(E),cmp)){
            steps = 0;
            break;
        }
        temp.x += dx;
        temp.y += dy;
        steps++;
    }
    return steps;
}
void solve(){
    qsort(a, n, sizeof(E), cmp);
    int maxi = 2, steps;
    for (int i = 0; i < n - 1; i++)
        for (int j = i + 1; j < n; j++){
            E temp;
            int dx = -a[i].x + a[j].x;
            int dy = -a[i].y + a[j].y;
            temp.x = a[i].x + maxi*dx;
            temp.y = a[i].y + maxi*dy;
            int px = a[i].x - dx;
            int py = a[i].y - dy;
            if (px>0&&px<=row&&py>0&&py<=col) continue;//如果上一个在稻田里面,那么说明这种情况已经枚举过了
            if (temp.x > row) break;//如果和显存的最大一样的时候就以及超出了稻田的范围那么就 不用再判断了,肯定小于目前的答案
            if (temp.y > col || temp.y <= 0) continue;//如果,y超出了界限肯定不是正确答案,注意,x的斜率一定是小于1的因为排序是按照x的非递减序列拍的,而y可以说是无序的,所以有斜率大于零与小于零的两种情况
            steps = search(a[j], dx, dy);
            if (steps > maxi) maxi = steps;

        }
    if (maxi == 2) maxi = 0;
    printf("%d\n", maxi);
    return;
}

int main(){
    cin >> row >> col;
    cin >> n;
    for (int i = 0; i < n; i++){
        cin >> a[i].x >> a[i].y;
    }
    solve();
    system("pause");
}