POJ-2479 Maximum sum (不连续的最大子段和)
Maximum sum
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 34398 | Accepted: 10661 |
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.
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
大致题意:给一个数列,求出数列中不想交的两个子段和,要求和最大。
思路:对于每个i来说,求出【0~i-1】的最大子段和以及【i~n-1】的最大子段和,再加起来,求最大的一个就行了。【0~i-1】的最大子段和从左向右扫描,【i~n-1】从右向左扫描即可。
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 int a[50001]; 5 int left[50001]; 6 int right[50001]; 7 int main() 8 { 9 int t; 10 scanf("%d",&t); 11 while(t--) 12 { 13 int n; 14 scanf("%d",&n); 15 for(int i=0;i<n;i++) 16 scanf("%d",&a[i]); 17 left[0]=a[0]; 18 for(int i=1;i<n;i++) 19 { 20 if(left[i-1]<0) 21 left[i]=a[i]; 22 else 23 left[i]=left[i-1]+a[i]; 24 } 25 for(int i=1;i<n;i++) 26 left[i]=max(left[i],left[i-1]); 27 right[n-1]=a[n-1]; 28 for(int j=n-2;j>=0;j--) 29 { 30 if(right[j+1]<0) 31 right[j]=a[j]; 32 else 33 right[j]=right[j+1]+a[j]; 34 } 35 for(int i=n-2;i>=0;i--) 36 right[i]=max(right[i+1],right[i]); 37 int res=-100000000; 38 for(int i=1;i<n;i++) 39 { 40 res=max(res,left[i-1]+right[i]); 41 } 42 printf("%d\n",res); 43 } 44 return 0; 45 }