PAT甲级 1046

题目 Shortest Distance

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.

解析

序列1到2为d1、序列2到3为d2依次类推，求两个序列之间的距离min(a到b,a到1+b到n)
如果用二位数组a[100005][100005]来存会超空间，两层循环会超时。所以用一个一维数组存储各左边的和，另一个一维数组存右边的和。

代码

#include<bits/stdc++.h>

#define INF 1<<29

using namespace std;

int n, data[100005];
int data_left[100005];
int data_right[100005];

void pat1046() {
    int l = 0, r = 0;
    data_left[1] = 0;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> data[i];
        l += data[i];
        data_left[i + 1] = l;
    }
    for (int i = n; i >= 1; --i) {
        r += data[i];
        data_right[i] = r;
    }
    int t = 0;
    cin >> t;
    for (int i = 0; i < t; ++i) {
        int a, b;
        cin >> a >> b;
        if (a > b)
            swap(a, b);
        int left = 0, right = 0;
        right = data_left[b] - data_left[a];
        if (a != 1)
            left += data_left[a];
        left += data_right[b];
        if (left < right) {
            cout << left << endl;
        } else {
            cout << right << endl;
        }
    }

}


int main() {
    pat1046();
    return 0;
}