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 S 1, S 2, S 3, S 4 … S x, … S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + … + S j (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(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + … + sum(i m, j m) maximal (i x ≤ i y ≤ j x or i x ≤ j y ≤ j x 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(i x, j x)(1 ≤ x ≤ m) instead. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S 1, S 2, S 3 … S n.
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
题目大意：给定n个数，和m个段。求在这n个数中求出m个子段，使得这些子段的和最大。就是区间求和的升级版
我的另一篇博文 最大区间和
思路：
//仿照区间求和得出状态转移方程
//j个数(j<=n) i段序列(i<=m)
//在i段序列中：对于第j个数来说他要么处在i段序列中最后一段的最后一个数
//要么是i段序列中最后的一段(第i段)，即可得出 状态转移方程
//dp[i][j]=max(dp[i][j-1]+A[j]，dp[i-1][k]+A[j]) (i-1<=k<=j-1)
//dp[i][j]就是j个数分成i段序列
//n太大所以用滚动数组实现上述状态转移方程。
#include<iostream>
#include<cstring>
#include<stdio.h>
#include<queue>
using namespace std;
const int MAX=1000009;
const int INF = 0x7fffffff;
int n,m;
int A[MAX],dp[MAX],tem[MAX];
int DP(int m,int n){
int ret;
for(int i=1 ; i<=m ; i++){//分成m段
ret = -INF;
for(int j=i ; j<=n ; j++){//n个数
//状态转移方程
dp[j] = max(dp[j-1]+A[j],tem[j-1]+A[j]);
//更新tem,使其为前j-1个数中的序列和最优解，并存下他
tem[j-1] = ret;
//更新ret其为前j个数中的序列和最优解
ret = max(ret,dp[j]);
}
}
return ret;
}
int main(void){
while(scanf("%d%d",&m,&n)!=EOF){
dp[0] = tem[0]=0;
for(int i=1 ; i<=n ; i++){
cin>>A[i];
tem[i]=0;
dp[i]=0;
}
cout<<DP(m,n)<<endl;
}
return 0;
}