HDU1024 DP的优化 最大M子段和问题

Max Sum Plus Plus

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

Total Submission(s): 31583    Accepted Submission(s): 11174
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[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.
只能说明自己还是太渣。。。

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=m;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，一起做过还是搞忘了

#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(
ad07
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。

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