poj1050

 1 //Accepted    200 KB    16 ms
 2 //dp
 3 //类比一维的情况
 4 //枚举i和j（i<=j），可以通过预处理在O(1)的时间求出a[k][i],a[k][i+1],...a[k][j]的和
 5 //然后就是纵向类比一维的情况
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <iostream>
 9 using namespace std;
10 const int imax_n = 105;
11 int a[imax_n][imax_n];
12 int d[imax_n];
13 int n;
14 int max(int a,int b)
15 {
16     return a>b?a:b;
17 }
18 void Dp()
19 {
20     for (int i=1;i<=n;i++)
21     {
22         for (int j=1;j<=n;j++)
23         a[i][j]+=a[i][j-1];
24     }
25     int ans=-1000000;
26     for (int i=1;i<=n;i++)
27     {
28         for (int j=i;j<=n;j++)
29         {
30             memset(d,0,sizeof(d));
31             d[1]=a[1][j]-a[1][i-1];
32             for (int k=2;k<=n;k++)
33             {
34                 d[k]=max(d[k-1]+a[k][j]-a[k][i-1],a[k][j]-a[k][i-1]);
35                 ans=max(d[k],ans);
36             }
37         }
38     }
39     printf("%d\n",ans);
40 }
41 int main()
42 {
43     while (scanf("%d",&n)!=EOF)
44     {
45         for (int i=1;i<=n;i++)
46         {
47             for (int j=1;j<=n;j++)
48             scanf("%d",&a[i][j]);
49         }
50         Dp();
51     }
52     return 0;
53 }

 1 //Accepted    244 KB    79 ms
 2 //枚举
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <iostream>
 6 using namespace std;
 7 const int imax_n = 105;
 8 int a[imax_n][imax_n];
 9 int sum[imax_n][imax_n];
10 int n;
11 void slove()
12 {
13     memset(sum,0,sizeof(sum));
14     for (int i=1;i<=n;i++)
15     for (int j=1;j<=n;j++)
16     sum[i][j]=a[i][j]+sum[i][j-1];
17     for (int i=1;i<=n;i++)
18     for (int j=1;j<=n;j++)
19     sum[i][j]+=sum[i-1][j];
20     int ans=-10000;
21     for (int i=1;i<=n;i++)
22     for (int j=1;j<=n;j++)
23     {
24         for (int k=i;k<=n;k++)
25         for (int l=j;l<=n;l++)
26         {
27             int temp=sum[k][l]-sum[i-1][l]-sum[k][j-1]+sum[i-1][j-1];
28             if (temp>ans) ans=temp;
29         }
30     }
31     printf("%d\n",ans);
32 }
33 int main()
34 {
35     while (scanf("%d",&n)!=EOF)
36     {
37         for (int i=1;i<=n;i++)
38         for (int j=1;j<=n;j++)
39         scanf("%d",&a[i][j]);
40         slove();
41     }
42     return 0;
43 }