HDU 1024(DP+滚动数组)

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 29920 Accepted Submission(s): 10505

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 … Sx, … Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + … + Sj (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(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx 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(ix, jx)(1 ≤ x ≤ m) instead. ^_^

Input

Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 … Sn.

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

Sample Output

6

8

Hint

Huge input, scanf and dynamic programming is recommended.

Author

JGShining（极光炫影）

Recommend

题意：求m个不相交子段的和的最大值

做法：

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

dp[i][j]表示以s[j]结尾的i个子段的最大和

结果是MLE

优化：滚动数组。当面状态只与i-1与i有关。我们可以开个2*n数组就够了

即:dp[1][j]=max(dp[1][j-1],dp[0][k])+s[j]

结果TLE

可以记录一下dp[0][k]的之前最大值来直接和dp[1][j-1]比较

即 dp[j]=max(dp[j-1],pre[j-1])+s[j]

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define maxn 1000010
#define inf 0x3f3f3f3f3f3f
using namespace std;
typedef long long ll;
ll n,m;
int s[maxn];
int dp[maxn];
int pre[maxn];
void init()
{
memset(dp,0,sizeof(dp));
memset(pre,0,sizeof(pre));
}
int main()
{
while(scanf("%I64d%64d",&m,&n)!=EOF)
{
init();
for(int i=1;i<=n;i++)
scanf("%d",&s[i]);
int tmp=-inf;
for(int i=1;i<=m;i++)
{
tmp=-inf;
for(int j=i;j<=n;j++)
{
dp[j]=max(dp[j-1],pre[j-1])+s[j];
pre[j-1]=tmp;
tmp=max(dp[j],tmp);
}
}
printf("%d\n",tmp);
}

return 0;
}