百练2479：Maximum sum

Maximum sum

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 41559 Accepted: 12981

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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好吧，我得承认，这道题的算法我并没有看懂。。连续最大子段和问题是基本算法，找了网上的算法：

int MaxSub (int a[])  {        int dp[N], max, i;        max = dp[0] = a[0];        for (i=1; i<N; i++)        {            if (dp[i-1] > 0)                dp[i] = dp[i-1] + a[i];            else                dp[i] = a[i];            if (dp[i] > max)                max = dp[i];        }        return max;    }  

dp[i]表示有a[i]参与的情况下某段序列的和。若前面的序列和dp[i-1]大于0，则把当前a[i]加到前面序列和dp[i-1]上，反之，则从a[i]处另起一段，赋给dp[i]。我就是这里不理解。为什么要根据dp[i-1]判断而不是a[i]呢？记下来吧，万一考呢。下面是这道题的算法：

//poj 2479 Maximum sum  //2013-05-01-17.26  #include <stdio.h>  #include <string.h>  #include <algorithm>  using namespace std;  const int maxn = 50005;  int dplift[maxn];  int dpright[maxn];  int a[maxn];    int main()  {      int t, n;      scanf("%d", &t);      while (t--)      {          scanf("%d", &n);          for (int i = 1; i <= n; i++)          {              scanf("%d", &a[i]);          }            dplift[1] = a[1];          for (int i = 2; i <= n; i++)          {              if (dplift[i-1] > 0)                  dplift[i] = dplift[i-1] + a[i];              else                  dplift[i] = a[i];          }          for (int i = 2; i <= n; i++)              dplift[i] = max(dplift[i], dplift[i-1]);            dpright[n] = a[n];          for (int i = n-1; i >= 1; i--)          {              if (dpright[i+1] > 0)                  dpright[i] = dpright[i+1] + a[i];              else                  dpright[i] = a[i];          }          for (int i = n-1; i >= 1; i--)              dpright[i] = max(dpright[i+1], dpright[i]);            int ans = dplift[1] + dpright[2];          for (int i = 2; i < n; i++)          {              ans = max(dplift[i]+dpright[i+1], ans);          }          printf("%d\n", ans);      }      return 0;  }  

用dplift[i]保存第1到第i个之间的最大子段和，dpright[i]保存第i到第n个之间的最大子段和，最终结果就是dplift[i]+dpright[i+1]中最大的一个。注意后一段要倒着求。

这是另一版代码;

#include <cstdio>#include <cstdlib>#include <cstring>#include <climits>const int MAXN = 50010;int an[MAXN];int d1[MAXN], d2[MAXN];int main(){    int cases;    scanf("%d", &cases);    while (cases--)    {        int n;        scanf("%d", &n);        for (int i = 1; i <= n; ++i)            scanf("%d", &an[i]);        int tmax = INT_MIN, temp = 0;        for (int i = 1; i <= n; ++i)        {            temp += an[i];            if (tmax < temp)                tmax = temp;            if (temp < 0)                temp = 0;            d1[i] = tmax;        }        tmax = INT_MIN, temp = 0;        for (int i = n; i >= 1; --i)        {            temp += an[i];            if (tmax < temp)                tmax = temp;            if (temp < 0)                temp = 0;            d2[i] = tmax;        }        int ans = INT_MIN;        for (int i = 1; i < n; ++i)            if (ans < d1[i] + d2[i+1])                ans = d1[i] + d2[i+1];        printf("%d\n", ans);    }    return 0;}

侵删。。。。

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坚持，胜利就在眼前~