Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15898    Accepted Submission(s): 5171

Problem Description

Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n). Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx is not allowed). But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^

Input

Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn. Process to the end of file.

Output

Output the maximal summation described above in one line.

Sample Input

1 3 1 2 3

2 6 -1 4 -2 3 -2 3

Sample Output

6

8

Hint

Huge input, scanf and dynamic programming is recommended.

Author

JGShining（极光炫影）

题意：

给定一段连续数字串，要你求出m个连续子串的最大值。

比如

下面这个样列

2  6 -1 4 -2 3 -2 3

给出一个长度为6的连续数字串，要你求出这个串中连续的m=2个子串的子串之和.......

aa

-1

4

-2

3

-2

3

-1

4

4

4

5

\

dp

-1

4

2

5

3

6

-1

3

3

7

7

\

dp

-1

3

2

7

5

8

``` 1 #include<iostream>
2 #include<string.h>
3 #include<stdio.h>
4 using namespace std;
5 int a[1000001],dp[1000001],max1[1000001];
6  int max(int x,int y){
7     return x>y?x:y;
8 }
9   int main(){
10     int i,j,n,m,temp;
11     while(scanf("%d%d",&m,&n)!=EOF)
12     {
13          dp[0]=0;
14         for(i=1;i<=n;i++)
15         {
16           scanf("%d",&a[i]);
17           dp[i]=0;
18           max1[i]=0;
19          }
20         max1[0]=0;
21         for(i=1;i<=m;i++){
22             temp=-0x3f3f3f3f;
23             for(j=i;j<=n;j++){
24                 dp[j]=max(dp[j-1]+a[j],max1[j-1]+a[j]);
25                 max1[j-1]=temp;
26                 temp=max(temp,dp[j]);
27             }
28         }
29         printf("%d\n",temp);
30     }
31     return 0;
32   }```

优化后的代码：

``` 1 /*hdu 1024 @coder Gxjun*/
2 #include<iostream>
3 #include<cstdio>
4 #include<cstring>
5 #include<cstdlib>
6 using namespace std;
7 const int maxn=1000005;
8 int aa[maxn],dp[maxn],maxc[maxn];
9 int max(int a,int b){
10   return a>b?a:b;
11 }
12 int main()
13 {
14     int n,m,i,j,temp;
15     while(scanf("%d%d",&m,&n)!=EOF){
16         memset(maxc,0,sizeof(int)*(n+1));
17         memset(dp,0,sizeof(int)*(n+1));
18        for(i=1;i<=n;i++)
19           scanf("%d",&aa[i]);
20        for(i=1;i<=m;i++){
21           temp=-0x3f3f3f3f;
22        for(j=i;j<=n;j++){
23           dp[j]=max(dp[j-1],maxc[j-1])+aa[j];
24           maxc[j-1]=temp;
25           temp=max(temp,dp[j]);
26           }
27         }
28         printf("%d\n",temp);
29     }
30 }```

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