hdu 1024 Max Sum Plus Plus

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13213    Accepted Submission(s): 4355

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 ... 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(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + sum(i_m, j_m) maximal (i_x ≤ i_y ≤ 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 S₁, S₂, S₃ ... 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 recommended.

 

 

Author

JGShining（极光炫影）

题意大致就是说给n个数，求出m个子串和的最大值。

此题确实有点难想，但是找到啦状态转移方程就比较好想，假设dp[i][j]为前j个数n个串的和的最大值，所以dp[i][j]=max{dp[i][j-1]+a[j],dp[i-1][j-1]+a[i]}(a[i]为第i个数）。

#include<stdio.h>
#include<string.h>
#define Max(a,b) (a>b?a:b)
#define size 1000100
#define INF 1<<30
int dp[2][size];//这儿只用啦一个dp[0],d[1]表示，可以减少内存的浪费，关键是这题我们只需要知道在j-1个数的i和i-1个子串和的最大值
int num[size];
int main()
{
    int n,m,i,j,max;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
            dp[1][i]=dp[0][i]=0;
        }
        for(i=1;i<=m;i++)//表示有i个子串
        {
            max=-INF;//令一个很小的数，好把在第i个子串是第一次的值保存下来
            for(j=i;j<=n;j++)
            {
                dp[1][j]=Max(dp[1][j-1],dp[0][j-1])+num[j];//此时的dp[1][j-1]代表前j-1个数有i个子串和的最大值，dp[0][j-1]代表前j-1个数i-1个子串的最大值（注意：i-1个子串中得最后一个子串都是以a[j-1]结尾的），并且还有一种特殊情况就是j=i时，因为也可用这方程表示就用啦一个方程表示；
                dp[0][j-1]=max;//记录当前状态（即i个子串）前j-1个数的最大值，以便在求前j个i+1子串是用
                if(max<dp[1][j])
                {
                    max=dp[1][j];//比较得出前j个数i个子串和的最大值
                }
            }
        }
        printf("%d\n",max);
    }
    return 0;
}