PAT 1046 Shortest Distance （20 分）

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 [3,1$0^5$​ ]), 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 ($\le 10^4$), 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 $10^7$

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Output Specification:
For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

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Sample Input:

5 1 2 4 14 9
3
1 3
2 5
4 1

Sample Output:

3
10
7

解析

这题有些难
exit的数量最大为$10^5$。而查询的数量最大为$10^4$。而Time Limit是100ms。所以用普通方法肯定会超时。必须优化。

exit是构成一个圆环的，如上图：所以可以顺时针走，也可以逆时针走。题目要求的是最短路径。
其中：顺时针走的距离+逆时针走的距离 = 走一圈的距离。所以这里可以优化。
举个例子：从2走到4，可以变为从1走到4的距离减去从1走到2的距离。所以我们可以存储从1到其他结点的距离。这样：又优化了一次。
经过上面的改良：可以从O(N)变为O(1).这样就不会超时了。
Code:

#include<vector>
#include<iostream>
#include<utility>
using namespace std;
int main()
{
	int N, M,Circule=0;
	cin >> N;
	vector<int> A(N + 1, 0),distance(N+1,0);
	for (int i = 1; i <= N; i++) {
		cin >> A[i];
		Circule += A[i];
		distance[i] = distance[i-1]+A[i];
	}
	cin >> M;
	pair<int, int> A2B;
	while (M--) {
		cin >> A2B.first >> A2B.second;
		if (A2B.first > A2B.second)
			swap(A2B.first, A2B.second);
		int shortest1 = distance[A2B.second-1]-distance[A2B.first-1];
		int shortest2 = Circule-shortest1;
		cout<<(shortest1>shortest2?shortest2:shortest1)<<endl;
	}
}