Live2d Test Env

HDU1024 最大M子段和问题 (单调队列优化)

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31583    Accepted Submission(s): 11174


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
Hint
Huge input, scanf and dynamic programming is recommended.
 
不加优化:
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<memory.h>
using namespace std;
int dp[2][1000010],a[1000010];
int main()
{
    int n,m,j,i,k,Max;
    while(~scanf("%d%d",&m,&n)){
        Max=0;
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++) scanf("%d",&a[i]);
        for(i=1;i<=m;i++) 
         for(j=i+1;j<=n;j++){
                dp[i%2][j]=dp[i%2][j-1]+a[j];
                for(k=i-1;k<=j-1;k++)
                 if(dp[(i-1)%2][k]+a[j]>dp[i%2][j]) dp[i%2][j]=dp[(i-1)%2][k]+a[j];
                if(i==m&&dp[i%2][j]>Max) Max=dp[i%2][j];
         }
         printf("%d\n",Max);
    }
    return 0;
}

 

然后发现k的范围【i-1,j-1】之间可以直接记录一个Maxp
emmmmm,以前做过还是搞忘了
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<memory.h>
using namespace std;
int dp[2][1000010],a[1000010];
int main()
{
    int n,m,j,i,k,Max,Maxp;
    while(~scanf("%d%d",&m,&n)){
        Max=-1000000001;
        for(i=1;i<=n;i++) scanf("%d",&a[i]);
        for(i=1;i<=n;i++) dp[0][i]=dp[1][i]=0;
        
         for(i=1;i<=m;i++) {
           Maxp=dp[(i-1)%2][i-1];
           dp[i%2][i]=dp[(i-1)%2][i-1]+a[i];
           for(j=i+1;j<=n-m+i;j++){
                if(dp[(i-1)%2][j-1]>Maxp) Maxp=dp[(i-1)%2][j-1];
                dp[i%2][j]=dp[i%2][j-1]+a[j];
                if(Maxp+a[j]>dp[i%2][j]) dp[i%2][j]=Maxp+a[j];
         }
        }
         for(i=m;i<=n;i++)
           if(dp[m%2][i]>Max) Max=dp[m%2][i];
         printf("%d\n",Max);
    }
    return 0;
}

 

至于此题的数据范围,呵呵,不存在的。






 
posted @ 2017-09-27 21:06  nimphy  阅读(301)  评论(0编辑  收藏  举报