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 }
原文地址：https://www.cnblogs.com/GetcharZp/p/9589797.html