A - Max Sum Plus Plus （好题&&dp）
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. ^_^
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. ^_^
InputEach 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.
OutputOutput 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.
题意：给你n个数，让你从中取出m个子段使其和最大
题解：dp，dp[i][j]表示到a[j]包括a[j]从中去i段的最大值
所以dp[i][j]就分为两种情况：a[j]取或者不取
dp[i%2][k]=max(dp[i%2][k-1],w[k]);
用w[i]记录一定取的情况，又分为两种情况：
1、a[k]作为第i段
2、a[k]加到之前的最大段那里
w[k]=max(dp[(i-1)%2][k-1],w[k-1])+sum[k]-sum[k-1];
初值： dp[0][i]
if(i==k）dp[i%2][k]=w[k]=sum[k];
具体看代码：
#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int,int> PII;
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
//head
#define INF 0x3f3f3f3f
#define N 1000005
int n,m;
int w[N];
int dp[2][N];
int sum[N];
int a[N];
int main()
{
ios_base::sync_with_stdio(0); cin.tie(0);
while(cin>>m>>n){
sum[0]=0;
for(int i=1;i<=n;i++)
{
cin>>a[i];
sum[i]=sum[i-1]+a[i];
dp[0][i]=0;
}
for(int i=1;i<=m;i++)
{
for(int k=i;k<=n;k++)
{
if(i==k)
dp[i%2][k]=w[k]=sum[k];//从k个数中取k段的最大值是前k个数的和
else
{
w[k]=max(dp[(i-1)%2][k-1],w[k-1])+sum[k]-sum[k-1];//这是一定要取的情况，分为两种：1、a[k]作为第i段，2、a[k]加到之前的最大段那里
dp[i%2][k]=max(dp[i%2][k-1],w[k]);//a[k]取或者不取
}
}
}
cout<<dp[m%2][n]<<endl;
}
return 0;
}
参考博客：http://blog.sina.com.cn/s/blog_677a3eb30100jxqa.html
https://blog.csdn.net/lishuhuakai/article/details/8067474
