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 0x3f3f3f3f3f3fusing 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;}