題目

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

思路

這個逼題目好題目，首先狀態方程很好寫：\(dp[i][j]=max (dp[i-1][j],dp[t][j-1])+a[i]\)

程式碼實現

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
const int inf=0x3f3f3f3f;
void check_min (int &a,int b) {a=min (a,b);}
void check_max (int &a,int b) {a=max (a,b);}
int n,m;
int a[maxn],last_max [maxn];
int dp[maxn];

int main () {
    while (~scanf ("%d%d",&m,&n)) {
        for (int i=1;i<=n;i++) scanf ("%d",&a[i]);
        memset (dp,0,sizeof (dp));
        memset (last_max,0,sizeof (last_max));
        int temp;
        for (int i=1;i<=m;i++) {
            temp=-1e9;
            for (int j=i;j<=n;j++) {
                dp[j]=max (dp[j-1],last_max[j-1])+a[j];
                last_max[j-1]=temp;
                check_max (temp,dp[j]);
            } 
        }
        printf ("%d\n",temp);
    }

    return 0;
}