AtCoder 5288-Balanced Pat【多维dp】 难度：***

题意：

We have a grid with H horizontal rows and W vertical columns. Let (i,j) denote the square at the i-th row from the top and the j-th column from the left.

The square (i, j) has two numbers A_{ij} and B_{ij} written on it.

First, for each square, Takahashi paints one of the written numbers red and the other blue.

Then, he travels from the square (1, 1) to the square (H, W). In one move, he can move from a square (i, j) to the square (i+1, j) or the square (i, j+1). He must not leave the grid.

Let the unbalancedness be the absolute difference of the sum of red numbers and the sum of blue numbers written on the squares along Takahashi’s path, including the squares (1, 1) and (H, W).

Takahashi wants to make the unbalancedness as small as possible by appropriately painting the grid and traveling on it.

Find the minimum unbalancedness possible.

Constraints
2 \leq H \leq 80
2 \leq W \leq 80
0 \leq A_{ij} \leq 80
0 \leq B_{ij} \leq 80
All values in input are integers.
Input
Input is given from Standard Input in the following format:

H W
A_{11} A_{12} \ldots A_{1W}
:
A_{H1} A_{H2} \ldots A_{HW}
B_{11} B_{12} \ldots B_{1W}
:
B_{H1} B_{H2} \ldots B_{HW}
Output
Print the minimum unbalancedness possible.

Sample Input 1
Copy
2 2
1 2
3 4
3 4
2 1
Sample Output 1
Copy
0
By painting the grid and traveling on it as shown in the figure below, the sum of red numbers and the sum of blue numbers are 3+3+1=7 and 1+2+4=7, respectively, for the unbalancedness of 0.

Figure

Sample Input 2
Copy
2 3
1 10 80
80 10 1
1 2 3
4 5 6
Sample Output 2
Copy
2

题解：

多维dp题，开一个三维数组dp[x][y][k]表示截止到坐标(x，y)点上是否能达成两组数差值为k，因为表示的是能不能，所以这个dp开成bool型就行。整张地图最大80×80，因此从左上角走到右下角最多走159个格子，考虑极限情况每个格子的两个数差值都是80，那么k的取值范围应该在[-159×80,159×80]。考虑到负数不能做下标的原因，可以整体增大159×80。

代码：

#include<stdio.h>
#include<math.h>
int a[85][85],b[85][85];
bool dp[82][82][25445];  //一共有80*159*2=25440种取值，从-12720~12720 
int main()
{
	int m,n,derta,max,min;
	scanf("%d%d",&m,&n);
	for(int i=0;i<m;i++)for(int j=0;j<n;j++)scanf("%d",&a[i][j]);
	for(int i=0;i<m;i++)for(int j=0;j<n;j++)scanf("%d",&b[i][j]);
	derta=b[0][0]-a[0][0];
	max=12720+fabs(derta);
	min=12720-fabs(derta);
	dp[0][0][derta+12720]=true;
	dp[0][0][-derta+12720]=true;
	for(int i=0;i<m;i++)
	{
		for(int j=0;j<n;j++)
		{
			for(int k=(min<80?0:min-80);k<=(max>25366?25440:max+80);k++)
			{
				if(j>0)if(dp[i][j-1][k-(b[i][j]-a[i][j])])dp[i][j][k]=true;
				if(i>0)if(dp[i-1][j][k-(b[i][j]-a[i][j])])dp[i][j][k]=true;
				if(j>0)if(dp[i][j-1][k-(a[i][j]-b[i][j])])dp[i][j][k]=true;
				if(i>0)if(dp[i-1][j][k-(a[i][j]-b[i][j])])dp[i][j][k]=true;
				if(dp[i][j][k]==true&&k<min)min=k;
				if(dp[i][j][k]==true&&k>max)max=k;
			}
		}
	}
	int ans=25441,temp;
	for(int i=0;i<25441;i++)
	{
		if(dp[m-1][n-1][i]==true)if(fabs(i-12720)<ans)ans=fabs(i-12720);
		if(i>12720&&i>ans)break;
	}
	printf("%d\n",ans);
}