HDU1024 最大和子序列增強版

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5360    Accepted Submission(s): 1773

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

 

 

最大和子序列的進化版 , 要求從一序列中取出若干段 , 使得這幾段的和最大 .

設 dp[i][j] 爲前 j 個數字分成 i 段的最大和 .

轉移方程爲 :

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

其表達的意義就兩個不同的決策 : 前者表示與 j-1 所在的一段合併成一段 , 後者表示以

a[j] 爲首開始第 i 段 .

下面的代碼實現的時候用了滾動數組節約空間 , 可以參考一下 .

 

代碼如下 :

#include<stdio.h> #include<string.h> #define inf -2100000000 #define MAXN 1000005 int a[MAXN],dp[MAXN],tdp[2][MAXN],sum[MAXN]; inline int max (int a,int b) { return a>b?a:b; } int main() { int m,n,i,j,k,res; while (scanf("%d%d",&m,&n)!=EOF) { sum[0]=0; for (i=1; i<=n; ++i) { scanf("%d",&a[i]); sum[i]=sum[i-1]+a[i]; } memset(dp,0,sizeof(dp)); for (i=0;i<=n;++i) tdp[0][i]=0; for (i=0;i<=n;++i) tdp[1][i]=inf; k=0; for (i=1; i<=m; ++i) { tdp[1-k][i-1]=inf; for (j=i; j<=n; ++j) { if (i==j) dp[j]=sum[j]; else { dp[j]=max(dp[j-1],tdp[k][j-1])+a[j]; } tdp[1-k][j]=max(tdp[1-k][j-1],dp[j]); } k=1-k; //tdp[0]和tdp[1]交替使用 } res=inf; for (i=m; i<=n; ++i) { if (dp[i]>res) res=dp[i]; } printf("%d/n",res); } return 0; }