HDU1024 最大和子序列增强版

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 "MaxSum" problem. To be a brave ACMer, we always challenge ourselves to moredifficult 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 afunction 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 jwhich 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 haveto 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

 

 

最大和子序列的进化版,要求从一序列中取出若干段,使得这几段的和最大.

设dp[i][j]为前j个数字分成i段的最大和.

转移方程为:

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

其表达的意义就两个不同的决策:前者表示与j-1所在的一段合并成一段,后者表示以

a[j]为首开始第i段.

下面的代码实现的时候用了滚动数组节约空间,可以参考一下.

 

代码如下:

#include<stdio.h>#include<string.h>#define inf -2100000000#define MAXN 1000005int 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;}