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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）
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1074 1025 1081 1080 1160 大意：主要是降低复杂度，问给你m个数要可以分成n块连续的数，问最大和是多少
定义dp[i][j]表示当前这个数为i，已经分了j块
可以转来的方向 1.块数不动多一个数 2.块数动了多一个数
动态转移方程 dp[i][j] = max(dp[i-1][j]+a[i],max(dp[1][j-1].....dp[i-1][j-1])+a[i]))
bing神代码好流弊。。定义一个数组记录前面一个块的前j-1中最大的值
那么最后答案就是这个max2（分成n组时1到j中最大的值
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
int dp[maxn];
int a[maxn],max1[maxn];
int main()
{
int n,m;
int max2;
while(~scanf("%d%d",&n,&m)){
for(int i = 1; i <= m ;i++)
scanf("%d",&a[i]);
memset(dp,0,sizeof(dp));
memset(max1,0,sizeof(max1));
for(int i = 1; i <= n; i++){
max2 = -inf;
for(int j = i; j <= m; j++){
dp[j] = max(dp[j-1]+a[j],max1[j-1]+a[j]);
max1[j-1] = max2;
max2 = max(max2,dp[j]);
}
}
printf("%d\n",max2);
}
return 0;
}
HDU1024——DP——Max Sum Plus Plus
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原文地址：http://www.cnblogs.com/zero-begin/p/4523184.html
