HDOJ 题目1024 Max Sum Plus Plus（动态规划，不想交子段最大和）

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17234    Accepted Submission(s): 5674

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
Hint
Huge input, scanf and dynamic programming is recommended.
 

 

Author

JGShining（极光炫影）

 

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思路：http://blog.csdn.net/jdplus/article/details/19974647

设状态为 cur[i,j]，表示前 j 项分为 i 段的最大和，且第 i 段必须包含 data[j]，则状态转移方程如下：
cur[i,j] = max{cur[i,j − 1] + data[j],max{cur[i − 1,t] + data[j]}}, 其中i ≤ j ≤ n,i − 1 ≤ t < j
target = max{cur[m,j]}, 其中m ≤ j ≤ n
分为两种情况：
• 情况一，data[j] 包含在第 i 段之中，cur[i,j − 1] + data[j]。
• 情况二，data[j] 独立划分成为一段，max{cur[i − 1,t] + data[j]}。
观察上述两种情况可知 cur[i,j] 的值只和 cur[i,j-1] 和 cur[i-1,t] 这两个值相关，因此不需要二维数组，
可以用滚动数组，只需要两个一维数组，用 cur[j] 表示现阶段的最大值，即 cur[i,j − 1] + data[j]，用
pre[j] 表示上一阶段的最大值，即 max{cur[i − 1,t] + data[j]}。

ac代码

#include<stdio.h>#include<string.h>#define INF 0xfffffff#define max(a,b) (a>b?a:b)int dp[1000100],a[1000100],pre[1000100];int main(){int n,m;while(scanf("%d%d",&m,&n)!=EOF){int i,ans,max,j;for(i=1;i<=n;i++){scanf("%d",&a[i]);}memset(dp,0,sizeof(dp));memset(pre,0,sizeof(pre));for(i=1;i<=m;i++){max=-1*INF;for(j=i;j<=n;j++){dp[j]=max(dp[j-1]+a[j],pre[j-1]+a[j]);pre[j-1]=max;;//这两条语句不能写反了,pre[j-1]表示的是上一个状态中i...j-1的最大值，max更新之后表示的i...j的最大值，if(dp[j]>max)max=dp[j];}}printf("%d\n",max);}}