[Two-segment continuous non-duplicate sequence and maximum] dynamic planning

Maximum subsequence

Timelimit: 1 second memorylimit: 32 megabyte

Totalsubmit: 156 accepted: 42

Description

Given a sequence composed of N integers, integers are positive and negative, and two consecutive subsequences that do not overlap are found so that the integers and the largest of them are obtained. Both sub-sequences can be empty.

Input

Multiple Input groups. The first action N in each group indicates the length of the sequence. The second action n integers indicates the input sequence.
0 <n <= 1,000,000

Output

For each input group, only one integer is returned, indicating the maximum sum.

Sample Input

9
185-580-889 701 964-878 353-761 608

Sample output

2273

Hint

For example, the input sequence is (701 964) and (608), and the Integer Range is)

 

 

#include <iostream>#include <cstdio>using namespace std;const int INF=1000005;int s[INF],lt[INF],rt[INF],dp[INF];int main(){    //freopen("in.txt","r",stdin);    int i,n;    while(cin >> n)    {        for (i=1;i<=n;i++)scanf("%d",&s[i]);        dp[n+1]=dp[0]=-INF;        lt[0]=rt[n+1]=-INF;        for (i=1;i<=n;i++)//正向        {            dp[i] = max(dp[i-1]+s[i],s[i]);        }         for (i=1;i<=n;i++)         {             lt[i] = max(dp[i],lt[i-1]);//             lt[i] = dp[i];         }        for (i=n;i>=1;i--)  //逆向        {            dp[i] = max(dp[i+1]+s[i],s[i]);        }         for (i=n;i>=1;i--)         {             rt[i] = max(dp[i],rt[i+1]);//                rt[i] = dp[i];         }         int sum=-INF;//枚举        for (i=1;i<=n;i++)        {            sum = max(sum,lt[i]+rt[i+1]);        }        if(sum<=0)        cout <<‘0‘ <<endl;        else printf("%d\n",sum);    }    return 0;}

 

[Two-segment continuous non-duplicate sequence and maximum] dynamic planning