poj-2593 Max Sequence

Max Sequence

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 14335		Accepted: 6012

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

POJ Monthly--2005.08.28,Li Haoyuan

 

 

题意：求一个序列中两个不相交的连续子段的最大和

 /*代码一：  -----TLE
#include <cstdio>
#include <iostream>
#include <cstring>

using namespace std;

int num[100001], dp[2][100001];
//dp[i][j] 表示 前 j 项中选择 i 个连续子段所得的最大和，且第 i 个连续子段是以 num[j]结尾的 
int main()
{
    int n;
    while(scanf("%d", &n), n)
    {
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; ++i)
        {
            scanf("%d", &num[i]);
            dp[1][i] = num[i];
        }
        for(int j = 1; j <= 2; ++j)
        {
            for(int i = 1; i <= n; ++i)
            {
                for(int t = 1; t < i-1; ++t)
                    dp[j][i] = max(dp[j][i-1], dp[j-1][t]) + num[i];
            }
        }
        printf("%d
", dp[2][n]);
    }
    return 0;
}
*/
//代码二：--------AC 
#include <cstdio>
#include <iostream>
#include <cstring>

using namespace std;

int num[100001], dp[100001];
// dp[i] 存储前 i 项的最大字段和 
int main()
{
    int n, max, sum, ans;
    while(scanf("%d", &n), n)
    {
        sum = 0;
        max = -0x3fffffff;
        //memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; ++i)
        {
            scanf("%d", &num[i]);
            sum += num[i];
            if(sum > max)
                max = sum;
            dp[i] = max;
            if(sum < 0)
                sum = 0;
        }
        dp[0] = ans = max = -0x3fffffff;
        sum = 0;
        for(int i = n; i > 0; --i) // 逆向遍历 
        {
            sum += num[i];
            if(sum > max)
                max = sum;
            if(ans < max+dp[i-1])
                ans = max+dp[i-1];
            if(sum < 0)
                sum = 0;
        }
        printf("%d
", ans);
    }
    return 0;
}