HDU 1024 Max Sum Plus Plus [动态规划+m子段和的最大值]

Max Sum Plus Plus

Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22262    Accepted Submission(s): 7484
 
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

[题意]:输入一个m,n分别表示成m组,一共有n个数即将n个数分成m组,m组的和加起来得到最大值并输出。

[分析]:

状态dp[i][j]表示前j个数分成i组的最大值。

动态转移方程:dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][k])+a[j]) (0<k<j)

dp[i][j-1]+a[j]表示的是前j-1分成i组,第j个必须放在前一组里面。

max( dp[i-1][k] ) + a[j] )表示的前(0<k<j)分成i-1组,第j个单独分成一组。

但是题目的数据量比较到,时间复杂度为n^3,n<=1000000,显然会超时,继续优化。

max( dp[i-1][k] ) 就是上一组 0....j-1 的最大值。我们可以在每次计算dp[i][j]的时候记录下前j个
的最大值 用数组保存下来 ,这样时间复杂度为 n^2。

[代码]:

 /*
输入一个m,n分别表示成m组,一共有n个数
即将n个数分成m组,
m组的和加起来得到最大值并输出。
*/
#include <bits/stdc++.h>
using namespace std;
const int N=;
#define INF 0x7fffffff int a[N+];
int dp[N+],Max[N+]; int main()
{
int n,m,maxs;
while(~scanf("%d%d",&m,&n))
{
for(int i=;i<=n;i++)
{
scanf("%d",&a[i]);
}
memset(dp,,sizeof(dp));
memset(Max,,sizeof(Max)); for(int i=;i<=m;i++)
{
maxs=-INF;
for(int j=i;j<=n;j++)
{
dp[j]=max(dp[j-]+a[j], Max[j-]+a[j]);
Max[j-]=maxs;
maxs=max(maxs, dp[j]);
}
}
printf("%d\n",maxs);
}
}

线性DP

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