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[ACM] POJ 2479 Maximum sum (动态规划求不相交的两段子段和的最大值）

2024-07-16 01:27:09   220人阅读

Maximum sum
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 33363 Accepted: 10330

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

1

10
1 -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


解题思路：

题意要求为给定一个数字序列，找出两段不相交的子段，使这两个子段的和最大，求出这个最大值。

dp[i]表示 从位置1到i 之间的最大子段和，正向求一遍。然后逆向求最大子段和，比如逆向求出当前位置i的最大字段和为sum,那么 ans= max( ans,dp[i-1]+sum)， ans即为答案。

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int maxn=50010;
const int inf=-0x7fffffff;
int dp[maxn];
int num[maxn];
int t,n;

void DP()//正向求最大子段和
{
    memset(dp,0,sizeof(dp));
    int sum=inf,b=inf;
    for(int i=1;i<=n;i++)
    {
        if(b>0)
            b+=num[i];
        else
            b=num[i];
        if(b>sum)
        {
            sum=b;
            dp[i]=sum;
        }
    }
}

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&num[i]);
        DP();
        
        int ans=inf,b=0,sum=inf;//逆向求n到i最大字段和，与正向的最大字段和相加，求出最大值
        for(int i=n;i>1;i--)
        {
            if(b>0)
                b+=num[i];
            else
                b=num[i];
            if(b>sum)
                sum=b;
            if(sum+dp[i-1]>ans)
                ans=sum+dp[i-1];
        }
        printf("%d\n",ans);

    }
    return 0;
}



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