http://acm.hdu.edu.cn/showproblem.php?pid=1024
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. ^_^
题意就是求你n个数字m段和的最大值
用动态规划求dp[n][m] 表示用前m个数划分为n块的最大和
转移方程为 dp[n][m]=max(dp[n][m-1]+a[m],max(dp[n-1][t])+a[m]),其中dp[n][m-1]+a[m]表示第m个在前i块中 后面表示的是在自己独立变成一个块（1<=t<=m-1）
那么问题来了如果 用这种方法 复杂度为（n^3）因此我们要化简一下，求max(dp[n-1][t])的时候重复求了一些过程，因此只需要改成 设一个变量M 表示前1到m-1的最大值 M=max(M,dp[i-1][j-1]);
但是空间上不够，因此用一个滚动数组 dp[2][MAN]；在求M的用到了i-1;
完整代码：
#include <stdlib.h> #include<iostream> #include<algorithm> //#include<bits/stdc++.h> #include<cstdio> #include<cstring> #include<cmath> #define ll long long const ll INF=0x3f3f3f3f; #define mod 1000000007 #define mem(a,b) memset(a,b,sizeof(a)) //__builtin_popcount using namespace std; //priority_queue const ll MAX=1000000; int Max[MAX+10]; int dp[2][MAX+10]; int a[MAX+10]; int main() { int n,m; while(cin>>m>>n) { for(int i=1; i<=n; i++) scanf("%d",&a[i]); mem(Max,0); mem(dp,0); int M; for(int i=1; i<=m; i++) { M=-0x3f3f3f3f; for(int j=i; j<=n; j++) { M=max(M,dp[(i-1)%2][j-1]); if(i!=j) dp[i%2][j]=max(dp[i%2][j-1]+a[j],M+a[j]); else dp[i%2][j]=M+a[j]; } } M=-0x3f3f3f3f; for(int i=m; i<=n; i++) M=max(M,dp[m%2][i]); cout<<M<<endl; } }
原文：https://www.cnblogs.com/zxz666/p/11190614.html
