立志用最少的代碼做最高效的表達

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PAT甲級最優題解——>傳送門

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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

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題意：給定N，后接N個數，數i為第i個點到第i+1個點的距離。
輸入M，后接M組數。輸出從x到y的最短距離。

最初的思路：對于每組數，從前以及向后遍歷，分別得到兩個值，輸出最小值即可。

分析一下時間復雜度：極限數據為十萬個點，一萬組數據。若處理每組數據的規模為O(n)， 最后的計算量為10w*1w=10e。而一百毫秒大約可以運行三十萬組數據， 顯然超時

那么就要求我們把求最短距離的復雜度控制在常數級O(1)

改進思想：一次性累加、保存所有點到點1的距離，對于任意兩點間距離的計算，只需相減即可。具體邏輯請閱讀代碼體會。

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#include<bits/stdc++.h> using namespace std; int main() {ios::sync_with_stdio(false);int n; cin >> n;vector<int>dis(n+1);int sum = 0, left, right, cnt;for(int i = 1; i <= n; i++) {int temp; cin >> temp;sum += temp;dis[i] = sum;} // for(int i = 1; i <= n; i++) { // cout << dis[i] << ' '; // }cin >> cnt;for(int i = 0; i < cnt; i++) {cin >> left >> right;if(left > right) swap(left, right); int temp = dis[right-1]-dis[left-1]; //順時針這條路 cout << min(temp, sum-temp) << '\n'; //sum-temp是逆時針這條路 }return 0; }

耗時：

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求贊哦~ (?ω?)

總結

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