题目：
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27299 Accepted Submission(s):
9499
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.
与求最大连续子段略有不同，此题目要求求出m个连续子段的sum，且使得sum尽量大；
考虑状态转移方程：f{i,j}表示前j个元素分为i段的最大值，则有 f{i,j}=max{f{i,j-1},max{i-1,k}}+a[j],(i-1<=k<j);
两种决策：
-1-: 前j个元素分为i段切最后一段以a结尾；
-2-：aj独自占一段，剩下的i-1段选前k（i-1<=k<j)个元素的i-1段中的最大值
n为100w考虑使用滚动数组，pre[j]保存上一个i对应的j的的最大值，dp[j]为当前正在计算的i的j的最大值；
代码：
#include<bits/stdc++.h>
using namespace std;
const int inf=999999999;
int dp[1000005],a[1000005],pre[1000005];
int main()
{
int n,m,i,j,k,t;
while (cin>>m>>n){int maxn=-inf;t=0;
memset(dp,0,sizeof(dp));
memset(pre,0,sizeof(pre));
for (i=1;i<=n;i++) scanf("%d",&a[i]);
for (i=1;i<=m;i++){
maxn=-inf; //每一轮开始都将maxn初始化
for (j=i;j<=n;j++){
dp[j]=max(pre[j-1],dp[j-1])+a[j]; //先使用在更新上一轮的值
pre[j-1]=maxn; //此处很重要，不可写成下文注释的形式，因为一旦那样写下一个j计算时用到的j-1变成了本轮的最大值的意思，概念就变了
if (maxn<dp[j]) maxn=dp[j];
//pre[j]=maxn;
}
//for(int l=0;l<=n;l++) cout<<pre[l]<<" ";
//cout<<endl;
}
cout<<maxn<<endl;
}
return 0;
}
hdu 1024 最大M子段dp
原文：http://www.cnblogs.com/zzqc/p/6412848.html
