poj 2479 Maximum sum 两段不相交最大子段和

Maximum sum

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 31315 Accepted: 9595

Description

Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:

Your task is to calculate d(A).

Input

The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input. 
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.

Output

Print exactly one line for each test case. The line should contain the integer d(A).

Sample Input

1101 -1 2 2 3 -3 4 -4 5 -5

Sample Output

13

Hint

In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer. 

Huge input,scanf is recommended.

Source

POJ Contest,Author:Mathematica@ZSU

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题意很简单，分别从左和从右求最大子段和 然后枚举中间点即可  之和就是最大sum

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;const int INF=1e9;int a[55555];int dp[55555];int L[55555],R[55555];int solve(int a[],int n){    int i;    dp[0]=-INF;    for(i=1;i<=n;i++)        dp[i]=max(dp[i-1]+a[i],a[i]);    L[0]=-INF;    for(i=1;i<=n;i++)        L[i]=max(dp[i],L[i-1]);    dp[n+1]=-INF;    for(i=n;i>=1;i--)        dp[i]=max(dp[i+1]+a[i],a[i]);    R[n+1]=-INF;    for(i=n;i>=1;i--)        R[i]=max(dp[i],R[i+1]);    int ans=-INF;    for(i=1;i<=n;i++)        ans=max(ans,L[i]+R[i+1]);    return ans;}int main(){    int t;    cin>>t;    while(t--)    {        int n;        int i;        scanf("%d",&n);        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        printf("%d\n",solve(a,n));    }    return 0;}