pat 1046 Shortest Distance（20 分） (线段树)

1046 Shortest Distance（20 分）

The task is really simple: given N exits on a highway which forms a simple cycle, you are supposed to tell the shortest distance between any pair of exits.

Input Specification:

Each input file contains one test case. For each case, the first line contains an integer N (in [3,10​5​​]), followed by N integer distances D​1​​ D​2​​ ⋯ D​N​​, where D​i​​ is the distance between the i-th and the (i+1)-st exits, and D​N​​ is between the N-th and the 1st exits. All the numbers in a line are separated by a space. The second line gives a positive integer M (≤10​4​​), with M lines follow, each contains a pair of exit numbers, provided that the exits are numbered from 1 to N. It is guaranteed that the total round trip distance is no more than 10​7​​.

Output Specification:

For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

Sample Input:

5 1 2 4 14 9
3
1 3
2 5
4 1

Sample Output:

3
10
7
 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <map>
 6 #include <stack>
 7 #include <vector>
 8 #include <queue>
 9 #include <set>
10 #define LL long long
11 using namespace std;
12 const int MAX = 4e5 + 10;
13 
14 int N, D[MAX], pre[MAX], M, ans, a, b, ans2;
15 struct node
16 {
17     int L, R, val;
18 }P[MAX];
19 
20 void build(int dep, int l, int r)
21 {
22     P[dep].L = l, P[dep].R = r, P[dep].val = 0;
23     if (l == r)
24     {
25         pre[l] = dep;
26         return;
27     }
28     int mid = (l + r) >> 1;
29     build(dep << 1, l, mid);
30     build((dep << 1) + 1, mid + 1, r);
31 }
32 
33 void update(int r, int b)
34 {
35     P[r].val += b;
36     if (r == 1) return ;
37     update(r >> 1, b);
38 }
39 
40 void query(int dep, int l, int r)
41 {
42     if (P[dep].L == l && P[dep].R == r)
43     {
44         ans += P[dep].val;
45         return ;
46     }
47     int mid = (P[dep].L + P[dep].R) >> 1;
48     if (r <= mid) query(dep << 1, l, r);
49     else if (l > mid) query((dep << 1) + 1, l, r);
50     else
51     {
52         query(dep << 1, l, mid);
53         query((dep << 1) + 1, mid + 1, r);
54     }
55 }
56 
57 int main()
58 {
59 //    freopen("Date1.txt", "r", stdin);
60     scanf("%d", &N);
61     build(1, 1, N);
62     for (int i = 1; i <= N; ++ i)
63     {
64         scanf("%d", &D[i]);
65         update(pre[i], D[i]);
66     }
67 
68     scanf("%d", &M);
69     while (M --)
70     {
71         ans = 0;
72         scanf("%d%d", &a, &b);
73         if (b < a) swap(a, b);
74         if (a == b - 1) ans += D[a];
75         else query(1, a, b - 1);
76         ans2 = ans, ans = 0;
77 
78         if (a - 1 == 1) ans += D[1];
79         else if (a - 1 > 1) query(1, 1, a - 1);
80         if (b == N) ans += D[N];
81         else query(1, b, N);
82         printf("%d\n", min(ans, ans2));
83     }
84     return 0;
85 }

 

posted @ 2018-09-05 03:11  GetcharZp  阅读(142)  评论(0)  编辑  收藏  举报