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代码量大,不过效率不错,分享下我的代码。

Posted by S070703520 at 2009-07-26 02:04:36 on Problem 1088
先后定义了两个类,重载了三个运算符,为了提高效率还用了快排。只不过代码太多了。
#include<stdio.h>

class Place
{
public:
	int h;
	int row;
	int col;
	Place operator = (Place x)
	{
		h=x.h;
		row=x.row;
		col=x.col;
		return *this;
	}
	bool operator > (Place x)
	{
		return(h>x.h);
	}
	bool operator < (Place x)
	{
		return(h<x.h);
	}
};

class Location
{
public:
	int hh;
	int count;
};

void swap(Place &a,Place &b)
{
	Place temp;
	temp=a;
	a=b;
	b=temp;
}

int partition(Place *a,int p,int r)
{   //划分算法
	int i=p,j=r+1;Place x=a[p];
	//将<=x的元素交换到左边区域
	//将>=x的元素交换到右边区域
	while(true)
	{
		while(a[++i]<x&&i<r);
		while(a[--j]>x);
		if(i>=j)break;
		swap(a[i],a[j]);
	}
	swap(a[j],a[p]);
	return j;
}

void quicksort(Place *a,int p,int r)
{   //随机化的快速排序算法
	if(p<r)
	{
		int q=partition(a,p,r);
		quicksort(a,p,q-1);//对左半段排序
		quicksort(a,q+1,r);//对右半段排序
	}
}

int main()
{
	int i,j,r,k,c,n,max=0,tmax,tmp,flag;
	Place pla[10003];
	Location loc[103][103];
	scanf("%d%d",&r,&c);
	n=r*c;
	k=0;
	for(i=0;i<=r+1;i++)
	{
		for(j=0;j<=c+1;j++)
		{
			if(i==0||i==r+1||j==0||j==c+1)
			{
				loc[i][j].hh=9999999;
			}
			else
			{
				scanf("%d",&loc[i][j].hh);
				pla[k].h=loc[i][j].hh;
				pla[k].row=i;
				pla[k++].col=j;
			}
		}
	}
	quicksort(pla,0,n-1);
	for(i=0;i<n;i++)
	{
		tmax=0;
		flag=0;
		if(loc[pla[i].row-1][pla[i].col].hh<loc[pla[i].row][pla[i].col].hh)
		{//上
			flag=1;
			tmp=loc[pla[i].row-1][pla[i].col].count+1;
			tmax=tmax>tmp?tmax:tmp;
		}
		if(loc[pla[i].row+1][pla[i].col].hh<loc[pla[i].row][pla[i].col].hh)
		{//下
			flag=1;
			tmp=loc[pla[i].row+1][pla[i].col].count+1;
			tmax=tmax>tmp?tmax:tmp;
		}
		if(loc[pla[i].row][pla[i].col-1].hh<loc[pla[i].row][pla[i].col].hh)
		{//左
			flag=1;
			tmp=loc[pla[i].row][pla[i].col-1].count+1;
			tmax=tmax>tmp?tmax:tmp;
		}
		if(loc[pla[i].row][pla[i].col+1].hh<loc[pla[i].row][pla[i].col].hh)
		{//右
			flag=1;
			tmp=loc[pla[i].row][pla[i].col+1].count+1;
			tmax=tmax>tmp?tmax:tmp;
		}
		if(flag==0)
		{
			tmax=loc[pla[i].row][pla[i].col].count=1;
		}
		loc[pla[i].row][pla[i].col].count=tmax;
		max=max>tmax?max:tmax;
	}
	printf("%d\n",max);
	return 0;
}


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