P2258 [NOIP2014 普及组] 子矩阵

题目

题目

思路

暴力枚举每一列情况，然后设 f x , y f_{x,y} fx,y​为前y行选x个的最优解， d x d_x dx​为x列所需代价 e x , y e_{x,y} ex,y​为x，y2行相邻的代价，则有：
f i , j = m i n ( f i − 1 , k + d j + e k , j ) ( 1 < = i < = c , i < = j < = m , 0 < = k < j ) f_{i,j}=min(f_{i-1,k}+d_j+e_{k,j})(1<=i<=c,i<=j<=m,\color{red}{0<=k<j}) fi,j​=min(fi−1,k​+dj​+ek,j​)(1<=i<=c,i<=j<=m,0<=k<j)
code：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring> 
using namespace std;
int n,m,r,c,a[20][20],b[20],d[20][20],e[20][20],f[20],ans=1<<30;
void dp()
{
	memset(b,0,sizeof(b));
	memset(e,0,sizeof(e));
	memset(d,127,sizeof(d));
	for (int i=1;i<=m;i++) for (int j=1;j<r;j++) b[i]+=abs(a[f[j]][i]-a[f[j+1]][i]);
	for (int i=1;i<=m;i++) for (int j=i+1;j<=m;j++) for (int k=1;k<=r;k++) e[i][j]+=abs(a[f[k]][i]-a[f[k]][j]);
	d[0][0]=0;
	for (int i=1;i<=c;i++)
	{
		for (int j=i;j<=m;j++)
		{
			for (int k=0;k<j;k++) d[i][j]=min(d[i][j],d[i-1][k]+b[j]+e[k][j]);
			if (i==c) ans=min(ans,d[c][j]);
		}
	}
	return;
}
void dfs(int x,int y)
{
	if (y>r)
	{
		dp();
		return;
	}
	if (x>n) return;
	dfs(x+1,y);
	f[y]=x;
	dfs(x+1,y+1);
	return;
}
signed main()
{
	cin>>n>>m>>r>>c;
	for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) cin>>a[i][j];
	dfs(1,1);
	cout<<ans;
	return 0;
}

标签：NOIP2014,20,P2258,int,memset,矩阵,dfs,sizeof,include
来源： https://blog.csdn.net/weixin_49843717/article/details/116393745