hdoj 1024 Max Sum Plus Plus 【简单dp】



Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21561    Accepted Submission(s): 7227

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 32 6 -1 4 -2 3 -2 3

 

Sample Output

68
分析：
最大区域子段和问题
代码：
#include<cstdio>#include<cstring>#include<cstdlib>#include<queue>#include<algorithm>#include<stack>#define mem(x,y) memset(x,y,sizeof(x))#define max(a,b) a>b?a:b#define min(a,b) a<b?a:b#define inf 0x80000000using namespace std;const int N=1e6+10;int pp[N],dp[N],bb[N];int main(){    int n,m,i,j,k,mmax;    while(scanf("%d%d",&n,&m)!=EOF)    {        for(i=1;i<=m;i++)            scanf("%d",&pp[i]);        for(i=0;i<=m;i++)            dp[i]=0,bb[i]=0;        for(i=1;i<=n;i++)        {            mmax=inf;            for(j=i;j<=m;j++)            {                if(dp[j-1]>bb[j-1])                    dp[j]=dp[j-1]+pp[j];                else dp[j]=bb[j-1]+pp[j];                bb[j-1]=mmax;                if(mmax<dp[j])                    mmax=dp[j];            }            bb[j-1]=mmax;            //printf("%d\n",bb[j-1]);        }        printf("%d\n",mmax);    }    return 0;}