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

 #include <iostream>
 #include <algorithm>
 #include <cstdio>
 #include <cstring>
 #include <map>
 #include <stack>
 #include <vector>
 #include <queue>
 #include <set>
 #define LL long long
 using namespace std;
 const int MAX = 4e5 + ;

 int N, D[MAX], pre[MAX], M, ans, a, b, ans2;
 struct node
 {
     int L, R, val;
 }P[MAX];

 void build(int dep, int l, int r)
 {
     P[dep].L = l, P[dep].R = r, P[dep].val = ;
     if (l == r)
     {
         pre[l] = dep;
         return;
     }
     int mid = (l + r) >> ;
     build(dep << , l, mid);
     build((dep << ) + , mid + , r);
 }

 void update(int r, int b)
 {
     P[r].val += b;
     if (r == ) return ;
     update(r >> , b);
 }

 void query(int dep, int l, int r)
 {
     if (P[dep].L == l && P[dep].R == r)
     {
         ans += P[dep].val;
         return ;
     }
     int mid = (P[dep].L + P[dep].R) >> ;
     if (r <= mid) query(dep << , l, r);
     else if (l > mid) query((dep << ) + , l, r);
     else
     {
         query(dep << , l, mid);
         query((dep << ) + , mid + , r);
     }
 }

 int main()
 {
 //    freopen("Date1.txt", "r", stdin);
     scanf("%d", &N);
     build(, , N);
     for (int i = ; i <= N; ++ i)
     {
         scanf("%d", &D[i]);
         update(pre[i], D[i]);
     }

     scanf("%d", &M);
     while (M --)
     {
         ans = ;
         scanf("%d%d", &a, &b);
         if (b < a) swap(a, b);
         if (a == b - ) ans += D[a];
         else query(, a, b - );
         ans2 = ans, ans = ;

         if (a -  == ) ans += D[];
         else if (a -  > ) query(, , a - );
         if (b == N) ans += D[N];
         else query(, b, N);
         printf("%d\n", min(ans, ans2));
     }
     return ;
 }