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 ≤ iy ≤ 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, S2, 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:从n个数中取m段和最大
dp[i][j]表示从前i个数中取j段(第i个数一定取), 所以i可以单独一段,也可以和前面构成连续的一段。推出状态转移方程dp[i][j]=max(dp[i-1][j]+a[i],dp[k][j-1]+a[i]) ,k的范围为j-1到i-1
但是这道题n和m比较大,二维的状态显然不行,所以需要降维。
#include <bits/stdc++.h> using namespace std; int a[100005],pre[100005],dp[100005]; int main() { int n,m,k; ios::sync_with_stdio(false); while(cin>>m>>n) { for(int i=1;i<=n;i++) cin>>a[i]; memset(pre,0,sizeof(pre)); memset(dp,0,sizeof(dp)); int maxx; for(int i=1;i<=m;i++) //分成i段 { maxx=-99999999; for(int j=i;j<=n;j++) { dp[j]=max(dp[j-1],pre[j-1])+a[j]; //取到第j个元素时,可以分成i段的最大值 pre[j-1]=maxx; //用前j-1个元素分成i段的最大值 maxx=max(maxx,dp[j]); } } cout<<maxx<<endl; } }