POJ1054 The Troublesome Frog [dp]

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艰难的写上一篇，小学期太累了，，，很难坚持刷

#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;
int m,n;
int num;
struct Grid
{
	int x,y;
}grid[5555];
int cmp(const void* a,const void* b)
{
	Grid* one;
	Grid* two;
	one=(Grid* ) a;
	two=(Grid* ) b;
	return ((one->x)==(two->x))?(one->y)-(two->y):(one->x)-(two->x);
}
int getstep(Grid grs,int dx,int dy)
{
	int step=0;
	while(grs.x<=n&&grs.x>=1 &&grs.y>=1&&grs.y<=m)
	{
		
		if(!bsearch(&grs,grid,num,sizeof(Grid),cmp))
		{
			step=0;
			break;
		}
		else
		{
			grs.x+=dx,grs.y+=dy;
			step++;
		}
	}
	return step;
}
int main()
{
	 //ios::sync_with_stdio(false);
    //freopen("/home/rainto96/in.txt","r",stdin);
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		scanf("%d",&num);
		for(int i=0;i<=num;i++)
		{
			int xx,yy;
			scanf("%d%d",&xx,&yy);
			grid[i].x=yy,grid[i].y=xx;
		}
		qsort(grid,num,sizeof(Grid),cmp);
		int maxn=2;
		for(int i=0;i<num-1;i++)
		{
			for(int j=i+1;j<num;j++)
			{
                    int dx,dy,px,py;
                    dx=grid[j].x-grid[i].x;
                    dy=grid[j].y-grid[i].y;
                    px=grid[i].x-dx;
                    py=grid[i].y-dy;
                    if(py>=1&&py<=m&&px>=1&&px<=n)
                        continue;
                    if(grid[i].x+maxn*dx>n)
                        break;
                    py = grid[i].y + maxn * dy;  
            		  if ( py > m || py < 1)  
                			continue; 
                    int step=getstep(grid[i],dx,dy);
                    maxn=max(maxn,step);
			}
		}
		maxn=(maxn==2?0:maxn);
		printf("%d\n",maxn);
	}
	return 0;
}

学会了二维的二分，其实就一句话

这个问题看起来很复杂，其实目的很简单：帮助农民找到为害最大的青蛙。也就是要找到

POJ1054 The Troublesome Frog [dp],布布扣,bubuko.com

POJ1054 The Troublesome Frog [dp]

标签：style   http   使用   width   数据   os   

原文地址：http://blog.csdn.net/u011775691/article/details/36551419