hdu 1024 Max Sum Plus Plus(DP最大字段和)

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 11158 Accepted Submission(s): 3673



[align=left]Problem Description[/align]
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. ^_^

[align=left]Input[/align]
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.

[align=left]Output[/align]
Output the maximal summation described above in one line.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

6
8

Hint
Huge input, scanf and dynamic programming is recommended.

[align=left]Author[/align]
JGShining（极光炫影）

思路：s[i][j]为i段包含j的最大和。ma[i][j]为前j个分为i段的最大和。答案就是ma[m]

d[j]存的是数

s[i][j]=max(s[i][j-1]+d[j],ma[i-1][j-1]+d[j]);

ma[i][j]=max(s[i][k]) (i<=k<=j)

又因为题目数给的太大一百万。所以二维数组不好开。想想可以用一维。

s[j]为i段含j最大和，ma【j】为i段前j个最大和。

s[j]=max(s[j-1]+d[j],ma[j-1]+d[j])

ma[j]=max(s[k]) （i<=k<=j）

ma[j]要先用后更新。这样就不会有问题了。

#include<cstdio>
#include<cstring>
const int mm=1000010;
const int oo=1e9;
int d[mm],s[mm],ma[mm];
int n,m;
int main()
{
while(scanf("%d%d",&m,&n)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",&d[i]);
memset(s,0,sizeof(s));
memset(ma,0,sizeof(ma));
int _min,_max;
for(int i=1;i<=m;i++)
{ _max=-oo;int j;
for(j=i;j<=n;j++)
{if(s[j-1]<ma[j-1])
s[j]=ma[j-1]+d[j];
else s[j]=s[j-1]+d[j];
ma[j-1]=_max;
if(s[j]>_max)
_max=s[j];
}
ma[j-1]=_max;
}
printf("%d\n",_max);
}
}