C++学习（1）：最大子段和（多种解法）

问题：给定由n个数（可能为负数）组成的序列a1,a2,a3,...,an,求该序列子段和的最大值。

第一种解法：（最容易考虑的方法，将所有的子段一一相加，然后比较）

 1 #include<iostream>
 2 using namespace std;
 3 int Maxsum(int n,int *a,int &besti,int &bestj)
 4 {
 5     int sum = 0;
 6     for(int i=1;i<=n;i++)
 7     {
 8         int thissum = 0;
 9         for(int j=i;j<=n;j++)
10         {
11             thissum +=a[j];
12             if(thissum>sum)
13             {
14                 sum = thissum;
15                 besti = i;
16                 bestj = j;
17             }
18         }
19     }
20     return sum;
21 }
22 
23 int main()
24 {
25     int Case = 1;
26     int T;//共有多少组数
27     cin>>T;
28     while(T>0)
29     {
30         T--;
31         int sum;
32         int N = 9;//一组有多少个数据
33         cin>>N;
34         int* array = new int[N+1];
35         for(int i=1;i<=N;i++)
36         {
37             cin>>array[i];
38         }
39         int besti,bestj;
40         sum = Maxsum(N,array,besti,bestj);
41         cout<<"Case "<<Case<<":"<<endl;
42         cout<<sum<<" "<<besti<<" "<<bestj<<endl;
43 
44         if(T>0)
45         {
46             cout<<endl;
47         }
48         Case++;
49     }
50     return 0;
51 }

 第二种解法：（利用分治算法，将序列从中间分为两个部分a[1:n/2]和a[n\2+1:n]，分别求最大子段和，然后比较。）

  1 #include<iostream>
  2 using namespace std;
  3 int MaxSubSum(int *a,int left,int right,int &besti,int &bestj)
  4 {
  5     int sum = 0;
  6     if(left==right)
  7     {
  8         sum = a[left];
  9         besti  = bestj = left;
 10     }
 11     else
 12     {
 13         int center = (left+right)/2;
 14         int leftsum = MaxSubSum(a,left,center,besti,bestj);
 15         int rightsum = MaxSubSum(a,center+1,right,besti,bestj);
 16 
 17         int s1 = -10001;
 18         int lefts = 0;
 19         for(int i = center;i>=left;i--)
 20         {
 21             lefts +=a[i];
 22             if(lefts>=s1)
 23             {
 24                 s1 = lefts;
 25                 besti = i;
 26                 
 27             }
 28         }
 29 
 30         int s2 = -10001;
 31         int rights = 0;
 32         for(int j = center+1;j<=right;j++)
 33         {
 34             rights +=a[j];
 35             if(rights>s2)
 36             {
 37                 s2 = rights;
 38                 bestj = j;
 39             }
 40         }
 41 
 42         sum = s1+s2;
 43 
 44         if(sum<leftsum)
 45         {
 46             sum = MaxSubSum(a,left,center,besti,bestj);
 47         }
 48         if(sum<rightsum)
 49         { 
 50             sum = MaxSubSum(a,center+1,right,besti,bestj);
 51         }
 52     }
 53     return sum;
 54 }
 55 
 56 int MaxSum(int n,int *a,int &besti,int &bestj)
 57 {
 58     return MaxSubSum(a,1,n,besti,bestj);
 59 }
 60 
 61 int main()
 62 {
 63     int Case = 1;
 64     int T;//共有多少组数
 65     cin>>T;
 66     while(T>0)
 67     {
 68         bool flag = true;
 69         T--;
 70         int sum;
 71         int N;//一组有多少个数据
 72         cin>>N;
 73         int* array = new int[N+1];
 74         for(int i=1;i<=N;i++)
 75         {
 76             cin>>array[i];
 77         }
 78         int besti,bestj;
 79         //判断是否全部为负数
 80         for(int j = 1;j<=N;j++)
 81         {
 82             if(array[j]>0) flag = false;
 83         }
 84         if(!flag)
 85         {
 86             sum = MaxSum(N,array,besti,bestj);
 87             cout<<"Case "<<Case<<":"<<endl;
 88             cout<<sum<<" "<<besti<<" "<<bestj<<endl;
 89         }else
 90         {
 91             sum = array[1];
 92             besti = bestj = 1;
 93             cout<<"Case "<<Case<<":"<<endl;
 94             cout<<sum<<" "<<besti<<" "<<bestj<<endl;
 95         }
 96         
 97         if(T>0)
 98         {
 99             cout<<endl;
100         }
101         Case++;
102     }
103     return 0;
104 }

 

posted on 2017-08-15 08:59  xcxfury  阅读(762)  评论(0)  编辑  收藏  举报

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