PAT1007 

 1 /*分治法找最大子串*/
 2 #include<iostream>
 3 #include<vector>
 4 using namespace std;
 5 #define M -99999999
 6 
 7 struct Node
 8 {
 9     int low;
10     int high;
11     int sum;
12 };
13 
14 Node find_max_crossing_subsequence(vector<int> A, int low, int mid, int high)
15 {
16     int left_sum = M;
17     int max_left;
18     int sum = 0;
19     for(int i=mid; i>=low; --i)
20     {
21         sum+=A[i];
22         if(sum >= left_sum)
23         {
24             left_sum = sum;
25             max_left = i;
26         }
27     }
28     int right_sum = M;
29     int max_right;
30     sum = 0;
31     for(int i=mid+1; i <= high; ++i)
32     {
33         sum += A[i];
34         if(sum > right_sum)
35         {
36             right_sum = sum;
37             max_right = i;
38         }
39     }
40     Node n={max_left, max_right, left_sum+right_sum};
41     return n;
42 }
43 
44 Node find_max_subsequence(vector<int> A, int low, int high)
45 {
46     if(low == high)
47     {
48         Node n={low, low, A[low]};
49         return n;
50     }
51     else
52     {
53         int mid = (low + high)/2;
54         Node ln = find_max_subsequence(A, low, mid);
55         Node cn = find_max_crossing_subsequence(A, low, mid, high);
56         Node rn = find_max_subsequence(A, mid+1, high);
57         if(ln.sum >= cn.sum && ln.sum >= rn.sum)
58             return ln;
59         else if(cn.sum >= ln.sum && cn.sum >= rn.sum)
60             return cn;
61         else
62             return rn;
63         /*比较顺序决定了当最大子串不止一个时返回子串的位置*/
64     }
65 }
66 
67 int main()
68 {
69     int num;
70     while(cin>>num)
71     {
72         int minus_num(0);
73         vector<int> A(num,0);
74         for(int i=0; i < num; ++i)
75         {
76             cin>>A[i]; 
77             if(A[i] < 0)
78                 ++minus_num;
79         }
80         if(minus_num == num)
81             cout<<0<<" "<<A[0]<<" "<<A[num-1]<<endl;
82         else
83         {
84             Node n = find_max_subsequence(A, 0, num-1);
85             cout<<n.sum<<" "<<A[n.low]<<" "<<A[n.high]<<endl;
86         }
87     }
88     return 0;
89 }

 

posted @ 2013-10-01 23:09  coding_monkey  阅读(193)  评论(0)  编辑  收藏  举报