noip 2014 子矩阵

先枚举行再DP列。好题，详见代码

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <vector>
#include <iostream>
using namespace std;
const int N=20;
const int inf =0x3f3f3f3f;
int n,m,r,c,p[N][N],v[N],f[N],res=inf,t[N],t1[N][N];

int DP() {
    memset(t,0,sizeof t);
    memset(t1,0,sizeof t1);
    for (int i=1; i<=m; i++)//预处理同一列相邻行之间的差值
        for (int j=1; j<v[0]; j++) t[i]+=abs(p[v[j]][i]-p[v[j+1]][i]);
    for (int i=1; i<m; i++)//处理每一列之间的差值
        for (int j=i+1; j<=m; j++)
            for (int k=1; k<=v[0]; k++) t1[i][j]+=abs(p[v[k]][i]-p[v[k]][j]);
    for (int i=1; i<=m; i++) f[i]=t[i];
    for (int i=2; i<=c; i++)//DP主要部分
        for (int j=m; j>=i; j--) {
            f[j]=inf;
            for (int k=j-1; k>=i-1; k--) f[j]=min(f[j],f[k]+t1[k][j]);
            f[j]+=t[j];
        }
    int ans=inf;
    for (int i=c; i<=m; i++) ans=min(ans,f[i]);
    return ans;
}

void DFS(int i,int dep) {//枚举行数
    if (dep==r) {
        res=min(res,DP());
        return;
    }
    for (int j=i; j<=n-r+dep+1; j++) {
        v[++v[0]]=j;
        DFS(j+1,dep+1);
        v[v[0]--]=0;
    }
}

void work() {
    DFS(1,0);
    printf("%d
",res);
}

int main() {
    memset(v,0,sizeof(v));
    scanf("%d%d%d%d",&n,&m,&r,&c);
    for (int i=1; i<=n; i++)
        for (int j=1; j<=m; j++)
            scanf("%d",&p[i][j]);
    work();
    return 0;
}