[Kuangbin take you to fly] twelve thematic basis DP1 A - Max Sum Plus Plus HDU - 1024

A - Max Sum Plus Plus

HDU - 1024

Topic links: https://vjudge.net/contest/68966#problem/A

topic:

Now I think you got a AC on "Max Sum" Ignatius.L the problem. To be a brave ACMer, we always challenge their own problems more difficult. Now you're faced with a more difficult problem.

    A given sequence of consecutive numbers S 1, S 2, S 3 , S 4 ... S x, ... S n (1≤x≤n≤1,000,000, -32768≤Sx≤32767). We define the function sum (i, j) = S i + ... + S j (1≤i≤j≤n).

    Now given an integer m (m> 0), the task is to find you for m i and j, and so that they (i 1, j 1) + sum (i 2, j 2) + sum (i 3, j 3 ) + ... + sum (im, jm) maximal ( or do not allow ix≤iy≤jx ix≤jy≤jx).

    But I'm lazy, I do not want to write a special judging module, so you do not have to output m i and j, and the maximum output only the sum of (ix, jx) (1≤x ≤m ) opposite. ^ _ ^
Input
    each test will start two integers m and n, followed by n integers S 1, S 2, S 3 ... S n.
    Treatment to the end of the file.
Yield
    outputs the sum of the maximum in a row.
Sample input

    . 1 2. 3. 3. 1
    2. 6. 3 -2 -2 -1. 4. 3

sample output

    . 6
    . 8

Ideas: borrow two pictures of other blog:

Such as: dp [2] [4] = max (max (-1,4,2,), 2) + num [4] = 7;

//
// Created by hy on 2019/8/4.
//
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <set>
#include<math.h>
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
int dp[maxn],frontmax[maxn],num[maxn];
#define MAX 0x3f3f3f3f
int main()
{
    int n,m;
    while(~scanf("%d%d",&m,&n))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&num[i]);
        memset(dp,0,sizeof(dp));
        memset(frontmax,0,sizeof(frontmax));
        int temp;
        for(int i=1;i<=m;i++)
        {
            temp=MAX*(-1);
            for(int j = I; j <= n-; j ++ ) 
            { 
                DP [j] = max (frontmax [J- . 1 ], DP [J- . 1 ;]) + NUM [j] // maximum before DP [j] j memory and, frontmax j-1 previously stored maximum value 
                frontmax [J- . 1 ] = TEMP; 
                TEMP = max (TEMP, DP [J]); // TEMP updated value is maximum 
            } 
        } 
        the printf ( " % D \ n- " , TEMP); 
    } 
    return  0 ; 
}