hdu-------1081To The Max
To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7681 Accepted Submission(s): 3724
Problem Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle. As an example, the maximal sub-rectangle of the array: 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2 is in the lower left corner: 9 2 -4 1 -1 8 and has a sum of 15.
Input
The input consists of an N x N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N 2 integers separated by whitespace (spaces and newlines). These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].
Output
Output the sum of the maximal sub-rectangle.
Sample Input
4
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
Sample Output
15
Source
这道题个人觉得是到灰常好的题目，适合不同层次的人做，求的是最大连续子矩阵和....
方法一：
简单的模拟即可。首先求出每个阶段只和，然后得到这个状态，然后进行矩阵规划，求最大值即可
代码浅显易懂：
1 #include<cstdio>
2 #include<cstring>
3 #include<cstdlib>
4 #include<iostream>
5 using namespace std;
6 int arr[101][101];
7 int sum[101][101];
8 int main()
9 {
10 int nn,i,j,ans,temp;
11 // freopen("test.in","r",stdin);
12 while(scanf("%d",&nn)!=EOF)
13 {
14 ans=0;
15 memset(sum,0,sizeof(sum));
16 for(i=1;i<=nn;i++)
17 {
18 for(j=1;j<=nn;j++)
19 {
20 scanf("%d",&arr[i][j]);
21 sum[i][j]+=arr[i][j]+sum[i][j-1]; //求出每一层逐步之和
22 }
23 }
24 for(i=2;i<=nn;i++)
25 {
26 for(j=1;j<=nn;j++)
27 {
28 sum[i][j]+=sum[i-1][j]; //在和的基础上，逐步求出最大和
29 }
30 }
31 ans=0;
32 for(int rr=1;rr<=nn;rr++)
33 {
34 for(int cc=1;cc<=nn;cc++)
35 {
36 for(i=rr;i<=nn;i++)
37 {
38 for(j=cc;j<=nn;j++)
39 {
40 temp=sum[i][j]-sum[i][cc-1]-sum[rr-1][j]+sum[rr-1][cc-1];
41 if(ans<temp) ans=temp;
42 }
43 }
44 }
45 }
46 printf("%d\n",ans);
47 }
48 return 0;
49 }方法二：
采用状态压缩的方式进行DP求解最大值......!
代码：
1 /*基于一串连续数字进行求解的扩展*/
2 /*coder Gxjun 1081*/
3 #include<iostream>
4 #include<cstdio>
5 #include<cstring>
6 #include<cstdlib>
7 using namespace std;
8 const int nn=101;
9 int arr[nn][nn],sum[nn][nn];
10 int main()
11 {
12 int n,i,j,maxc,res,ans;
13 // freopen("test.in","r",stdin);
14 while(scanf("%d",&n)!=EOF)
15 {
16 memset(sum,0,sizeof(sum));
17 for(i=1;i<=n;i++)
18 for(j=1;j<=n;j++)
19 {
20 scanf("%d",&arr[i][j]);
21 sum[j][i]=sum[j][i-1]+arr[i][j];
22 }
23 ans=0;
24 for(i=1;i<=n;i++){
25
26 for(j=i;j<=n;j++)
27 {
28 res=maxc=0;
29 for(int k=1;k<=n;k++)
30 {
31 maxc+=sum[k][j]-sum[k][i-1];
32 if(maxc<=0) maxc=0;
33 else
34 if(maxc>res) res=maxc;
35 }
36 if(ans<res) ans=res;
37 }
38 }
39 printf("%d\n",ans);
40 }
41 return 0;
42 }