dp[ i ][ j ]表示从1～～ j 分了 i 段，并且以 a[ j ]结尾的最大长度。
dp[ i ][ j ] = max( dp[ i ][ j-1] , dp[ i-1 ] [ k ]) +a[ j ] 即a[ j ] 可以接在最后一段后面，也可以另起一段
Total Submission(s): 12622 Accepted Submission(s): 4159
#include <iostream>
dp[ i ][ j ] = max( dp[ i ][ j-1] , dp[ i-1 ] [ k ]) +a[ j ] 即a[ j ] 可以接在最后一段后面，也可以另起一段
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12622 Accepted Submission(s): 4159
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. ^_^
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.
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
2 6 -1 4 -2 3 -2 3
Sample Output
6
8
8
Hint
Huge input, scanf and dynamic programming is recommended.Author
JGShining（极光炫影）
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a[1000100];
int dp[2][1000100];
int sum[1000100];
int f[1000100];
int n,m;
int main()
{
while(scanf("%d%d",&m,&n)!=EOF)
{
memset(dp,0,sizeof(dp));
sum[0]=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a);
sum=sum[i-1]+a;
}
int ans=-99999999;
int cur=0;
for(int j=1;j<=m;j++)
{
for(int i=0;i<=j-1;i++)
dp[1-cur]=-999999999;
for(int i=j;i<=n;i++)
{
if(i==j) f=sum;
else f=max(dp[cur][i-1],f[i-1])+a;//如果用3重循环就超时了
dp[1-cur]=max(dp[1-cur][i-1],f
if(j==m)
ans=max(ans,f);
}
cur=1-cur;
}
printf("%d\n",ans);
}
return 0;
}