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 [3,105]), followed by N integer distances D1 D2 ⋯ DN, where Di is the distance between the i-th and the (i+1)-st exits, and DN 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 (≤104), 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 107.

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:

 #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 ;

 }

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