poj 2593&&poj2479(最大两子段和)

Max Sequence

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 16850		Accepted: 7054

Description

Give you N integers a1, a2 ... aN (|ai| <=1000, 1 <= i <= N).

You should output S.

Input

The input will consist of several test cases. For each test case, one integer N (2 <= N <= 100000) is given in the first line. Second line contains N integers. The input is terminated by a single line with N = 0.

Output

For each test of the input, print a line containing S.

Sample Input

5
-5 9 -5 11 20
0

Sample Output

40

Source

求解序列中两段不相交的子序列的最大和。
先正着扫描 1- n-1 区间求出每个段的最大值，然后反着扫描 n - 2这段区间求出每个段的最大值，然后枚举1 - n-1 每个段，得到最大值

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
const int N = 100005;

int a[N];
int dp[N],dp1[N]; ///dp[i]第 1-i段的最大和 , dp1[i] 第 i - n段的最大和
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n){
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        int sum=0,mx = -100000000; ///每个数都有可能是负数
        for(int i=1;i<n;i++){   ///因为题目中两段不能重合，所以不能枚举到n
            sum +=a[i];
            if(sum>mx) mx = sum;
            if(sum<0){
                sum = 0;
            }
            dp[i]=mx;
        }
        //for(int i=1;i<=n;i++) printf("%d ",dp[i]);
        sum = 0,mx = -100000000;
        for(int i=n;i>1;i--)  ///因为题目中两段不能重合，所以不能枚举到1
        {
            sum+=a[i];
            if(sum>mx) mx = sum;
            if(sum<0) sum = 0;
            dp1[i] =mx;
        }
        //for(int i=1;i<=n;i++) printf("%d ",dp1[i]);
        mx = -100000000;
        for(int i=1;i<n;i++){
            if(dp[i]+dp1[i+1]>mx){
                mx = dp[i]+dp1[i+1];
            }
        }
        printf("%d\n",mx);
    }
    return 0;
}