In Korea, the naughtiness of the cheonggaeguri, a small frog, is legendary. This is a well-deserved reputation, because the frogs jump through your rice paddy at night, flattening rice plants. In the morning, after noting which plants have been flattened, you want to identify the path of the frog which did the most damage. A frog always jumps through the paddy in a straight line, with every hop the same length:
Your rice paddy has plants arranged on the intersection points of a grid as shown in Figure-1, and the troublesome frogs hop completely through your paddy, starting outside the paddy on one side and ending outside the paddy on the other side as shown in Figure-2:
Many frogs can jump through the paddy, hopping from rice plant to rice plant. Every hop lands on a plant and flattens it, as in Figure-3. Note that some plants may be landed on by more than one frog during the night. Of course, you can not see the lines showing the paths of the frogs or any of their hops outside of your paddy ?for the situation in Figure-3, what you can see is shown in Figure-4:
From Figure-4, you can reconstruct all the possible paths which the frogs may have followed across your paddy. You are only interested in frogs which have landed on at least 3 of your rice plants in their voyage through the paddy. Such a path is said to be a frog path. In this case, that means that the three paths shown in Figure-3 are frog paths (there are also other possible frog paths). The vertical path down column 1 might have been a frog path with hop length 4 except there are only 2 plants flattened so we are not interested; and the diagonal path including the plants on row 2 col. 3, row 3 col. 4, and row 6 col. 7 has three flat plants but there is no regular hop length which could have spaced the hops in this way while still landing on at least 3 plants, and hence it is not a frog path. Note also that along the line a frog path follows there may be additional flattened plants which do not need to be landed on by that path (see the plant at (2, 6) on the horizontal path across row 2 in Figure-4), and in fact some flattened plants may not be explained by any frog path at all.
Your task is to write a program to determine the maximum number of landings in any single frog path (where the maximum is taken over all possible frog paths). In Figure-4 the answer is 7, obtained from the frog path across row 6.
Input
Your program is to read from standard input. The first line contains two integers R and C, respectively the number of rows and columns in your rice paddy, 1 <= R,C <= 5000. The second line contains the single integer N, the number of flattened rice plants, 3 <= N <= 5000. Each of the remaining N lines contains two integers, the row number (1 <= row number <= R) and the column number (1 <= column number <= C) of a flattened rice plant, separated by one blank. Each flattened plant is only listed once.
Output
Your program is to write to standard output. The output contains one line with a single integer, the number of plants flattened along a frog path which did the most damage if there exists at least one frog path, otherwise, 0.
Sample Input
6 7 14 2 1 6 6 4 2 2 5 2 6 2 7 3 4 6 1 6 2 2 3 6 3 6 4 6 5 6 7
Sample Output
7
在存在n个脚印的格子内，是青蛙进过的痕迹，每次青蛙通过格子的路径是直线的，求符合条件的最多点直线。
思路：按从左到右，从上到下的顺序排序脚印的坐标，然后枚举每个点，作为直线的第一个点，注意剪枝不可能或遍历过情况，具体看代码
#include<cstdio>
#include<stack>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstring>
#include<string>
#include<map>
#include<iostream>
#include<cmath>
using namespace std;
#define inf 0x3f3f3f3f
typedef long long ll;
const int N=5016;
const int nmax=11;
const double esp = 1e-9;
const double PI=3.1415926;
int r,c,n;
struct point
{
int x,y;
} p[N];
int vis[N][N];
bool cmp(point p1,point p2)
{
if(p1.y==p2.y)
return p1.x<p2.x;
return p1.y<p2.y;
}
int judge(int a,int b)
{
if(a>0&&a<=r&&b>0&&b<=c)
return 1;
return 0;
}
int solve(int xx,int yy,int dx,int dy)
{
int ans=0;
while(judge(xx,yy)) //在格子内
{
if(!vis[xx][yy])
{
//不存在脚印，即直线不存在
return 0;
}
ans++;
xx+=dx;
yy+=dy;
}
return ans;
}
int main()
{
while(scanf("%d%d",&r,&c)!=EOF)
{
scanf("%d",&n);
memset(vis,0,sizeof(vis));
for(int i=0; i<n; i++)
{
scanf("%d%d",&p[i].x,&p[i].y);
vis[p[i].x][p[i].y]=1;
}
sort(p,p+n,cmp);
int sum=2;
for(int i=0; i<n; i++) //以p[i]作为直线的第一个点
{
for(int j=i+1; j<n; j++) //以p[i]作为直线的第二个点
{
int dx=p[j].x-p[i].x;
int dy=p[j].y-p[i].y;
if(p[i].y+sum*dy>c)
/*如果以p[i]作为直线的第二个点的直线可能超过sum个点，那么第sum+1个点应该在格子内，
否则这种情况是不存在多于sum个点的直线，同样地，以p[i]为起点，该层循环后面的y值比p[j].y更大，
即每次跳的幅度更大了，那么存在多于sum个点的直线更加不可能了，所以直接退出该层循环
*/
break;
if(judge(p[i].x+sum*dx,p[i].y+sum*dy)==0)
continue; //再判断x方向是否越界
if(judge(p[i].x-dx,p[i].y-dy))
//p[i]前面有一个点在格子内，在前面已经计算过了
continue;
sum=max(sum,solve(p[i].x,p[i].y,dx,dy));
}
}
if(sum<3)
printf("0\n");
else
printf("%d\n",sum);
}
return 0;
}