超级玛丽游戏——洛谷P1000

最新推荐文章于 2023-10-28 20:35:19 发布

cys2004a

最新推荐文章于 2023-10-28 20:35:19 发布

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CC 4.0 BY-SA版权

文章标签： c++ dp 矩阵打表

本文链接：https://blog.youkuaiyun.com/weixin_42781330/article/details/100596723

其实，这道题并不难，我们这里使用~~动态规划~~来解决这道题，这就是传说中的超级玛丽用dp。当然，编者a+b确是也TLE了。在此与各位读者分享一下心得。

本蒟蒻实在是太菜了，导致这道题只想出了一种dp+矩阵的做法，时间复杂度为O(n^2m)，不过优化后可变成O(nm)的。

#include<bits/stdc++.h>
using namespace std;
int a[25][65];
int g[23][23];
int f[63][23];
int ans[63][23];
void mul(int x)
{
	for(int i=0;i<22;i++)
	{
		for(int k=0;k<22;k++)
		{
			ans[x][i]=max(ans[x][i],g[i][k]+f[x][k]);
		}
	}
}
int main()
{
	memset(g,-0x3F,sizeof g);
	for(int i=0;i<22;i++)
	{
		g[i][i]=0;
	}
	string s[23];
	s[1]= "00000000000000003333333300000000000000000000000000000000000000";
	s[2]= "00000000000000033333333333300000000000000000000000000000000000";
	s[3]= "00000000000000011112222120000000000000000000000000000000000000";
	s[4]= "00000000000001221112222211222200000000000000000000000000000000";
	s[5]= "00000000000001112222222111111000000000000001110000000000001110";
	s[6]= "00000000000000002222222222200000000000000012221000000000012221";
	s[7]= "00000000000000011311111110000000000000000012121000000000012121";
	s[8]= "00000000000011113333333111111000000000000012121000000000012121";
	s[9]= "00000000000222133323333231112222000000000012221000000000012221";
	s[10]="00000000000222233333333331122222000000000001110000000000001110";
	s[11]="00000000000222233330000333332222000000000000000000000000000000";
	s[12]="00000000000001111000000001111000000000000000000000000000000000";
	s[13]="00000000000111111000000001111110000000000000000000000000000000";
	s[14]="11111111111111111111111111111111111111111111111111111111111111";
	s[15]="12221222222121122212222221211222122222212114444444444444444441";
	s[16]="11111111111111111111111111111111111111111114444444444444444441";
	s[17]="12212222122221122122221222211221222212222111111111111111111111";
	s[18]="11111111111111111111111111111111111111111100001444444444410000";
	s[19]="12222212222221122222122222211222221222222100001444444444410000";
	s[20]="11111111111111111111111111111111111111111100001444444444410000";
	s[21]="12122122221221121221222212211212212222122100001444444444410000";
	s[22]="11111111111111111111111111111111111111111100001111111111110000";
	for(int i=1;i<=22;i++)
	{
		for(int j=0;j<62;j++)
		{
			a[i][j]=s[i][j]-'0';
		}
	}
	for(int j=0;j<62;j++)
	{
		for(int i=1;i<=22;i++)
		{
			f[j][i-1]=a[i][j];//√ 
		}
	}
	for(int j=0;j<62;j++)
	{
		mul(j);
	}
	for(int j=0;j<62;j++)
	{
		for(int i=1;i<=22;i++)
		{
			a[i][j]=ans[j][i-1];
		}
	}
	for(int i=1;i<=22;i++)
	{
		for(int j=0;j<62;j++)
		{
			if(a[i][j]==0)
			{
				cout<<" ";
			}
			else if(a[i][j]==1)
			{
				cout<<"#";
			}
			else if(a[i][j]==2)
			{
				cout<<".";
			}
			else if(a[i][j]==3)
			{
				cout<<"*";
			}
			else if(a[i][j]==4)
			{
				cout<<"-";
			}
		}
		cout<<endl;
	}
	return 0;
}

这是洛谷上AC的代码。

当然，我们可以想办法去处理更复杂一点的超级玛丽游戏的界面，比如1000行，1000列。这时我们可以分析一下g矩阵，发现其除对角线外全是-∞，所以，我们可以用O(n)处理对角线，即可。

O(nm)代码：

#include<bits/stdc++.h>
using namespace std;
int a[25][65];
int g[23][23];
int f[63][23];
int ans[63][23];
void mul(int x)
{
	for(int i=0;i<22;i++)
	{
//		for(int k=0;k<22;k++)
//		{
			ans[x][i]=max(ans[x][i],g[i][i]+f[x][i]);
//		}
	}
}
int main()
{
	memset(g,-0x3F,sizeof g);
	for(int i=0;i<22;i++)
	{
		g[i][i]=0;
	}
	string s[23];
	s[1]= "00000000000000003333333300000000000000000000000000000000000000";
	s[2]= "00000000000000033333333333300000000000000000000000000000000000";
	s[3]= "00000000000000011112222120000000000000000000000000000000000000";
	s[4]= "00000000000001221112222211222200000000000000000000000000000000";
	s[5]= "00000000000001112222222111111000000000000001110000000000001110";
	s[6]= "00000000000000002222222222200000000000000012221000000000012221";
	s[7]= "00000000000000011311111110000000000000000012121000000000012121";
	s[8]= "00000000000011113333333111111000000000000012121000000000012121";
	s[9]= "00000000000222133323333231112222000000000012221000000000012221";
	s[10]="00000000000222233333333331122222000000000001110000000000001110";
	s[11]="00000000000222233330000333332222000000000000000000000000000000";
	s[12]="00000000000001111000000001111000000000000000000000000000000000";
	s[13]="00000000000111111000000001111110000000000000000000000000000000";
	s[14]="11111111111111111111111111111111111111111111111111111111111111";
	s[15]="12221222222121122212222221211222122222212114444444444444444441";
	s[16]="11111111111111111111111111111111111111111114444444444444444441";
	s[17]="12212222122221122122221222211221222212222111111111111111111111";
	s[18]="11111111111111111111111111111111111111111100001444444444410000";
	s[19]="12222212222221122222122222211222221222222100001444444444410000";
	s[20]="11111111111111111111111111111111111111111100001444444444410000";
	s[21]="12122122221221121221222212211212212222122100001444444444410000";
	s[22]="11111111111111111111111111111111111111111100001111111111110000";
	for(int i=1;i<=22;i++)
	{
		for(int j=0;j<62;j++)
		{
			a[i][j]=s[i][j]-'0';
		}
	}
	for(int j=0;j<62;j++)
	{
		for(int i=1;i<=22;i++)
		{
			f[j][i-1]=a[i][j];//√ 
		}
	}
	for(int j=0;j<62;j++)
	{
		mul(j);
	}
	for(int j=0;j<62;j++)
	{
		for(int i=1;i<=22;i++)
		{
			a[i][j]=ans[j][i-1];
		}
	}
	for(int i=1;i<=22;i++)
	{
		for(int j=0;j<62;j++)
		{
			if(a[i][j]==0)
			{
				cout<<" ";
			}
			else if(a[i][j]==1)
			{
				cout<<"#";
			}
			else if(a[i][j]==2)
			{
				cout<<".";
			}
			else if(a[i][j]==3)
			{
				cout<<"*";
			}
			else if(a[i][j]==4)
			{
				cout<<"-";
			}
		}
		cout<<endl;
	}
	return 0;
}

我们可不可以进一步的优化呢？答案是肯定的。我们可以发现，矩阵的对角线上全是0，所以乘完之后相当于一个间接的赋值，而这样看来后面的一整段都是~~没有意义的~~所以我们可以直接将打出来的表还原成字符串的即可，时间复杂度O(1)，空间复杂度O(n*m).

以下是本蒟蒻的代码：

#include<iostream>
using namespace std;
int main()
{
	cout<<"                ********"<<endl;
	cout<<"               ************"<<endl;
	cout<<"               ####....#."<<endl;
	cout<<"             #..###.....##...."<<endl;
	cout<<"             ###.......######              ###            ###"<<endl;
	cout<<"                ...........               #...#          #...#"<<endl;
	cout<<"               ##*#######                 #.#.#          #.#.#"<<endl;
	cout<<"            ####*******######             #.#.#          #.#.#"<<endl;
	cout<<"           ...#***.****.*###....          #...#          #...#"<<endl;
	cout<<"           ....**********##.....           ###            ###"<<endl;
	cout<<"           ....****    *****...."<<endl;
	cout<<"             ####        ####"<<endl;
	cout<<"           ######        ######"<<endl;
	cout<<"##############################################################"<<endl;
	cout<<"#...#......#.##...#......#.##...#......#.##------------------#"<<endl;
	cout<<"###########################################------------------#"<<endl;
	cout<<"#..#....#....##..#....#....##..#....#....#####################"<<endl;
	cout<<"##########################################    #----------#"<<endl;
	cout<<"#.....#......##.....#......##.....#......#    #----------#"<<endl;
	cout<<"##########################################    #----------#"<<endl;
	cout<<"#.#..#....#..##.#..#....#..##.#..#....#..#    #----------#"<<endl;
	cout<<"##########################################    ############"<<endl;
	return 0;
}