noip 2014 子矩阵

先枚举行再DP列。好题，详见代码

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 #include 
 6 #include 
 7 using namespace std;
 8 const int N=20;
 9 const int inf =0x3f3f3f3f;
10 int n,m,r,c,p[N][N],v[N],f[N],res=inf,t[N],t1[N][N];
11 
12 int DP() {
13     memset(t,0,sizeof t);
14     memset(t1,0,sizeof t1);
15     for (int i=1; i<=m; i++)//预处理同一列相邻行之间的差值
16         for (int j=1; j0]; j++) t[i]+=abs(p[v[j]][i]-p[v[j+1]][i]);
17     for (int i=1; i//处理每一列之间的差值
18         for (int j=i+1; j<=m; j++)
19             for (int k=1; k<=v[0]; k++) t1[i][j]+=abs(p[v[k]][i]-p[v[k]][j]);
20     for (int i=1; i<=m; i++) f[i]=t[i];
21     for (int i=2; i<=c; i++)//DP主要部分
22         for (int j=m; j>=i; j--) {
23             f[j]=inf;
24             for (int k=j-1; k>=i-1; k--) f[j]=min(f[j],f[k]+t1[k][j]);
25             f[j]+=t[j];
26         }
27     int ans=inf;
28     for (int i=c; i<=m; i++) ans=min(ans,f[i]);
29     return ans;
30 }
31 
32 void DFS(int i,int dep) {//枚举行数
33     if (dep==r) {
34         res=min(res,DP());
35         return;
36     }
37     for (int j=i; j<=n-r+dep+1; j++) {
38         v[++v[0]]=j;
39         DFS(j+1,dep+1);
40         v[v[0]--]=0;
41     }
42 }
43 
44 void work() {
45     DFS(1,0);
46     printf("%d\n",res);
47 }
48 
49 int main() {
50     memset(v,0,sizeof(v));
51     scanf("%d%d%d%d",&n,&m,&r,&c);
52     for (int i=1; i<=n; i++)
53         for (int j=1; j<=m; j++)
54             scanf("%d",&p[i][j]);
55     work();
56     return 0;
57 }

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转载于:https://www.cnblogs.com/ITUPC/p/5360789.html