hdu1231 最大连续子序列

 /*线性时间算法*/
1
#include<iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <algorithm> 5 using namespace std; 6 #define NMax 10010 7 int A[NMax]; 8 9 int main() 10 { 11 int n = 0; 12 while (scanf("%d", &n), n) 13 { 14 for (int k = 0; k < n; ++k) 15 { 16 scanf("%d", &A[k]); 17 } 18 19 __int64 subSum = 0; 20 __int64 sum = -1; 21 int i = 0; 22 int j = 0; //和不小于0的连续子数组的和的最大下标 23 int nStart = 0; 24 int nEnd = 0; //最大和连续子数组的最大下标 25 for (int k = 0; k < n; ++k) 26 { 27 subSum += A[k]; 28 if (subSum < 0) 29 { 30 subSum = 0; 31 i = k + 1; 32 j = k + 1; 33 } 34 else 35 { 36 if (subSum > sum) 37 { 38 sum = subSum; 39 nStart = i; 40 nEnd = k; 41 } 42 } 43 } 44 45 if (sum == -1) 46 { 47 printf("%d %d %d ", 0, A[0], A[n - 1]); 48 } 49 else 50 { 51 printf("%lld %d %d ", sum, A[nStart], A[nEnd]); 52 } 53 } 54 return 0; 55 }
  1 /*分治法 子串要么在左边,要么在右边,要么包含中间那个元素*/
  2 #include<iostream>
  3 #include <stdio.h>
  4 #include <stdlib.h>
  5 #include <algorithm>
  6 using namespace std;
  7 #define  NMax 10010
  8 int A[NMax];
  9 void AcrossMidMaxSum(int& sum, int left, int mid, int right, int& nStart, int& nEnd)
 10 {
 11     int leftMaxSum = INT_MIN;
 12     int leftSum = 0;
 13     for (int i = mid; i >= left; --i)
 14     {
 15         leftSum += A[i];
 16         if (leftSum >= leftMaxSum)
 17         {
 18             leftMaxSum = leftSum;
 19             nStart = i;
 20         }
 21     }
 22 
 23     int rightMaxSum = INT_MIN;
 24     int rightSum = 0;
 25     for (int i = mid + 1; i <= right; ++i)
 26     {
 27         rightSum += A[i];
 28         if (rightSum > rightMaxSum)
 29         {
 30             rightMaxSum = rightSum;
 31             nEnd = i;
 32         }
 33     }
 34 
 35     sum = leftMaxSum + rightMaxSum;
 36 }
 37 
 38 void MaxSum(int& sum, int left, int right, int& nStart, int& nEnd)
 39 {
 40     if (left == right)
 41     {
 42         sum = A[left];
 43         nStart = left;
 44         nEnd = right;
 45         return;
 46     }
 47 
 48     int mid = (left + right) / 2;
 49 
 50     int leftSum = -1, nLeftStart, nLeftEnd;
 51     MaxSum(leftSum, left, mid, nLeftStart, nLeftEnd);
 52 
 53     int midSum = -1, nMidStart, nMidEnd;
 54     AcrossMidMaxSum(midSum, left, mid, right, nMidStart, nMidEnd);
 55 
 56     int rightSum = -1, nRightStart, nRightEnd;
 57     MaxSum(rightSum, mid + 1, right, nRightStart, nRightEnd);
 58 
 59     if (leftSum >= midSum && leftSum >= rightSum)
 60     {
 61         sum = leftSum;
 62         nStart = nLeftStart;
 63         nEnd = nLeftEnd;
 64     }
 65     else if (midSum >= leftSum && midSum >= rightSum)
 66     {
 67         sum = midSum;
 68         nStart = nMidStart;
 69         nEnd = nMidEnd;
 70     }
 71     else
 72     {
 73         sum = rightSum;
 74         nStart = nRightStart;
 75         nEnd = nRightEnd;
 76     }
 77 }
 78 
 79 void Test_1231_1()
 80 {
 81     int n = 0;
 82     while (scanf("%d", &n), n)
 83     {
 84         for (int k = 0; k < n; ++k)
 85         {
 86             scanf("%d", &A[k]);
 87         }
 88         int sum = -1, nStart, nEnd;
 89         MaxSum(sum, 0, n-1, nStart, nEnd);
 90         if (sum < 0)
 91         {
 92             printf("%d %d %d
", 0, A[0], A[n - 1]);
 93         }
 94         else
 95         {
 96             printf("%d %d %d
", sum, A[nStart], A[nEnd]);
 97         }
 98     }
 99 }
100 
101 
102 int main()
103 {
104     //Test_1231_0();
105     Test_1231_1();
106     return 0;
107 }
原文地址:https://www.cnblogs.com/heqinghui/p/9063969.html