HDU 5805 - NanoApe Loves Sequence （BestCoder Round #86）

先找相邻差值的最大，第二大，第三大

删去端点会减少一个值， 删去其余点会减少两个值，新增一个值，所以新增和现存的最大的值比较一下取最大即可

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 using namespace std;
 5 #define LL long long
 6 const int N = 100005;
 7 int t, n, p1, p2, p3;
 8 LL a[N];
 9 LL s1[N], s2[N];
10 LL sum;
11 int main()
12 {
13     scanf("%d", &t);
14     while(t--)
15     {
16         scanf("%d", &n);
17         for(int i = 1; i <= n; i++)
18             scanf("%I64d", &a[i]);
19         p1 = p2 = p3 = 0;
20         s1[0] = -1;
21         for(int i = 1; i < n; i++)
22         {
23             s1[i] = abs(a[i + 1] - a[i]);//相邻 
24             if(s1[p1] < s1[i] ) p1 = i;//最大 
25         }
26         for(int i = 1; i < n; i++)
27         {
28             if(i == p1) continue;
29             else if(s1[p2] < s1[i]) p2 = i;//第二大 
30         }
31         for(int i = 1; i < n; i++)
32         {
33             if(i == p1 || i == p2) continue;//第三大 
34             else if(s1[p3] < s1[i]) p3 = i;
35         }
36         for(int i = 2; i < n; i++)
37             s2[i] = abs(a[i+1] - a[i-1]);//去掉 i 点新增值 
38         
39         sum = 0;
40         if(p1 == 1) sum += s1[p2]; 
41         else sum += s1[p1];
42         for(int i = 2; i < n; i++)
43         {
44             if(p1 == i-1 || p1 == i) 
45             {
46                 if(p2 == i-1 || p2 == i) sum += max(s1[p3], s2[i]);
47                 else sum += max(s1[p2], s2[i]);
48             } 
49             else sum += max(s1[p1], s2[i]);
50         }
51         if(p1 == n - 1) sum += s1[p2]; 
52         else sum += s1[p1];
53         printf("%I64d\n",sum);
54     }
55 }

 

posted @ 2016-08-06 22:05  nicetomeetu  阅读(190)  评论(0)  编辑  收藏  举报