hdu1024 Max Sum Plus Plus (Dp)

Posted 2020-09-16 Dysania_l

hdu1024 Max Sum Plus Plus

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 S 1, S 2, S 3, S 4 ... S x, ... S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + ... + S j (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(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(i m, j m) maximal (i x ≤ iy ≤ j x or i x ≤ j y ≤ j x 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(i x, j x)(1 ≤ x ≤ m) instead. ^_^ 

Input

Each test case will begin with two integers m and n, followed by n integers S 1, S2, S 3 ... S n. 

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 recommende

 

解题思路：

首先定义各个数组，尽量易懂，

a[i] 储存输入的数据

now[j]，表示以第j个元素为结尾的i个子段的最大和，必须包含a[j]。

pre[j]，表示前j个元素i个子段的最大和，不一定包含a[j]。

dp[i][j]，表示前j个元素i个子段的最大和，包含a[j]。

原始状态转移方程：

dp[i][j]=max(dp[i][j-1]+a[j]，dp[i-1][k]+a[j])         (i-1<=k<=j-1)

有两种情况：

1.直接将第j个元素加在第i个子段之后。

2.第j个元素单独作为一个子段，那么前面必须是i-1个子段。

状态转移方程就是比较这两种情况较大值

 

实现代码：

#include<iostream>
 #include<algorithm>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 #define M 1000002
 int a[M],now[M],pre[M];
 int main()
 {
     int m,n,i,j,Max;
     while(scanf("%d%d",&m,&n)==2)
     {
         for(i=1;i<=n;i++)  
             scanf("%d",&a[i]);
         memset(now,0,sizeof(now));
         memset(pre,0,sizeof(pre));
         for(i=1;i<=m;i++)
         {
             Max = -999999999;  //此处max值尽量开大，max是和，开小了就WA,
             for(j=i;j<=n;j++)
             {
                 now[j]=max(now[j-1]+a[j],pre[j-1]+a[j]);
                //now[j]有两种来源，一种是直接在第i个子段之后添加a[j]
                //一种是是a[j]单独成为1个子段
                 pre[j-1]=Max;       //更新pre使得pre是前j-1个中最大子段和
                 if(now[j]>Max)  
                 Max=now[j];
             }
         }
         printf("%d\n",Max);
     }
     return 0;
 }