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): 16250    Accepted Submission(s): 5311

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

求区间的和。。DP题。。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<cmath>typedef __int64 LL;using namespace std;const int N=1000000+100;const int INF = 0x3f3f3f3f;LL a[N];LL b[N];LL dp[N];LL Max;int main(){    int m, n;    while(scanf("%d%d", &m, &n)==2)    {        b[0] = dp[0] = 0;        for(int i=1; i<=n; i++)        {            scanf("%I64d", a+i);            b[i] = dp[i] = 0;        }        for(int i=1, j; i<=m; i++)        {            Max = -INF;            for(j=i; j<=n; j++)            {                if(b[j-1]>dp[j-1])                    dp[j] = b[j-1] + a[j];                else                    dp[j]=dp[j-1]+a[j];                b[j-1] = Max;                if(dp[j]>Max) Max = dp[j];            }            b[j-1] = Max;        }        printf("%I64d\n",Max);    }    return 0;}