题干：

 

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

Hint

Huge input, scanf and dynamic programming is recommended.

 

解题报告：

   dp[i][j]  i个子段，并且以第j个数字结尾  的最大和，然后优化掉一维。复杂度O(n*m)

AC代码：

#include<bits/stdc++.h>
#define ll long long
using namespace std;

const int MAX = 100005;
const ll INF = 922337203685477580;
ll num[MAX] ,dp[MAX] ,mk[MAX];
int main () {
	int n,m;
	ll maxx;
	while(~scanf("%d%d",&m,&n)) {
		for(int i = 1 ; i <= n ; i++) scanf("%lld",&num[i]);
		memset(dp,0,sizeof(dp));
		memset(mk,0,sizeof(mk));
		for(int i = 1 ; i <= m ; i ++) {
			maxx = -INF;
			for(int j = i ; j <= n ; j ++) {
				if(i == j) dp[j] = mk[j-1] + num[j];
				else dp[j] = max(dp[j-1] ,mk[j-1]) + num[j];
				mk[j-1] = maxx;
				maxx = max(maxx,dp[j]);
			}
		}
		printf("%lld\n",maxx);
	}
	return 0;
}

TLE代码：（复杂度O(m* n^2)）

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
ll dp[505][505];
ll a[MAX];
//dp[i][j]  i个子段，并且以第j个数字结尾  的最大和
int main() 
{
	int n,m;
	while(cin>>m>>n) {
		for(int i = 1; i<=n; i++) scanf("%lld",a+i);
		memset(dp,0,sizeof dp);
		for(int i = 1; i<=m; i++) {
			for(int j = 1; j<=n; j++) {
				dp[i][j] = dp[i][j-1] + a[j]; 
				if(j<=i) continue;
				for(int k = 1; k<j; k++) {
					dp[i][j] = max(dp[i][j],dp[i-1][k] + a[j]);
				}
			}
		}
		ll ans = -99999999999;
		for(int i = 1; i<=n; i++) ans = max(ans,dp[m][i]);
		cout <<ans <<endl;
	}
	return 0 ;
}

优化一层循环：（但是空间上过不去。。。）

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
ll dp[505][505];
ll pre[505][505];
ll a[MAX];
//dp[i][j]  i个子段，并且以第j个数字结尾  的最大和
int main() 
{
	int n,m;
	while(~scanf("%d%d",&m,&n)) {
		for(int i = 1; i<=n; i++) scanf("%lld",a+i);
		memset(dp,0,sizeof dp);
		memset(pre,0,sizeof pre);
		for(int i = 1; i<=m; i++) {
			for(int j = 1; j<=n; j++) {
				dp[i][j] = dp[i][j-1] + a[j]; 
				if(j<=i) continue;
				dp[i][j] = max(dp[i][j],pre[i-1][j-1] + a[j]);
				pre[i][j] = max(pre[i][j-1],dp[i][j]);
			}
		}
		ll ans = -99999999999;
		for(int i = 1; i<=n; i++) ans = max(ans,dp[m][i]);
		printf("%lld\n",ans);
	}
	return 0 ;
}