Max Sum Plus Plus(動態規劃難題:m段子段和的最大值)
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38381 Accepted Submission(s): 13747
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
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
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
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.
Author
JGShining(極光炫影)
AC code:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+5;
int a[maxn];
//int dp[maxn][maxn];
int d[maxn];
int pre[maxn];
int main()
{
// freopen("D:\\in.txt","r",stdin);
int m,n,i,j,k;
while(scanf("%d%d",&m,&n)!=EOF)
{
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
// memset(dp,0,sizeof(dp));
// for(i=1;i<=n;i++)
// {
// for(j=1;j<=m;j++)
// {
// dp[i][j]=dp[i-1][j]+a[i];
// for(k=1;k<i;k++)
// {
// dp[i][j]=max(dp[i][j],dp[k][j-1]+a[i]);
// }
// }
// }
memset(d,0,sizeof(d));
memset(pre,0,sizeof(pre));
int nowans=0;
for(j=1;j<=m;j++)
{
nowans = -1e9;
for(i=j;i<=n;i++)
{
d[i]=max(d[i-1],pre[i-1])+a[i];
pre[i-1]=nowans;
nowans = max(nowans,d[i]);
}
}
printf("%d\n",nowans);
}
return 0;
}