动态规划 hdu 1024
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41033 Accepted Submission(s):
14763
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.
^_^
followed by n integers S1, S2,
S3 ... Sn.
Process to the end of
file.
line.
2 6 -1 4 -2 3 -2 3
8
Huge input, scanf and dynamic programming is recommended;
int dp[maxn],pre[maxn],arr[maxn];
int temp,n,m;
arr[ ]储存输入的数;
for(int k=;k<=m;k++)
{
temp=-inf;
for(int j=k;j<=n;j++)
{
dp[j]=max(dp[j-],pre[j-])+arr[j];
pre[j-]=temp;
temp=max(temp,dp[j]);
}
}
用temp来找出 j 个数取 k 组所得的组的最大和;同时把它记入在pre[ ]中,用于进行下次更新;
dp[j]=max(dp[j-1],pre[j-1])+arr[j];此时dp[j]和dp[j-1]有同样的组数,
dp[j]=dp[j-1]+arr[j]; // 表示把第j个数加入dp[j-1]的其中一组,能保持组数不变;
dp[j]=pre[j-1]+arr[j]; // 表示让第j个数独成一组,再加上比dp[j]少一组的组集中的最大的;
两者中选择大的组合方式;
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std; const int maxn=+;
const int inf=0x3f3f3f3f; int dp[maxn],pre[maxn],arr[maxn];
int temp,n,m; int main()
{
while(~scanf("%d%d",&m,&n))
{
for(int i=;i<=n;i++)
{
scanf("%d",&arr[i]);
}
memset(dp,,sizeof(dp));
memset(pre,,sizeof(pre));
for(int k=;k<=m;k++)
{
temp=-inf;
for(int j=k;j<=n;j++)
{
dp[j]=max(dp[j-],pre[j-])+arr[j];
pre[j-]=temp;
temp=max(temp,dp[j]);
}
}
printf("%d\n",temp);
}
return ;
}
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