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 + 10;

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 = 0;
    if (l == r)
    {
        pre[l] = dep;
        return;
    }
    int mid = (l + r) >> 1;
    build(dep << 1, l, mid);
    build((dep << 1) + 1, mid + 1, r);
}

void update(int r, int b)
{
    P[r].val += b;
    if (r == 1) return ;
    update(r >> 1, 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) >> 1;
    if (r <= mid) query(dep << 1, l, r);
    else if (l > mid) query((dep << 1) + 1, l, r);
    else
    {
        query(dep << 1, l, mid);
        query((dep << 1) + 1, mid + 1, r);
    }
}

int main()
{
//    freopen("Date1.txt", "r", stdin);
    scanf("%d", &N);
    build(1, 1, N);
    for (int i = 1; i <= N; ++ i)
    {
        scanf("%d", &D[i]);
        update(pre[i], D[i]);
    }

    scanf("%d", &M);
    while (M --)
    {
        ans = 0;
        scanf("%d%d", &a, &b);
        if (b < a) swap(a, b);
        if (a == b - 1) ans += D[a];
        else query(1, a, b - 1);
        ans2 = ans, ans = 0;

        if (a - 1 == 1) ans += D[1];
        else if (a - 1 > 1) query(1, 1, a - 1);
        if (b == N) ans += D[N];
        else query(1, b, N);
        printf("%d
", min(ans, ans2));
    }
    return 0;
}