PAT 1046. Shortest Distance (20)

1046. Shortest Distance (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 ... 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 (<=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 14 931 32 54 1

Sample Output:

3107

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这道题和之前一道用二分法做的题有点类似，单纯的迭代计算会有一个点运行超时。不过这道题不用二分法，只要在输入数据的时候记录下每个点到第一个点的距离（顺时针），最后能记录下圈的周长。然后求任意两个点X,Y的最短距离就是Y到1的距离减去X到1的距离，记为clockwise，然后逆时针的距离就是周长减去clockwise，然后两个取其小就是答案了。代码如下： 

#include <iostream>#include <vector>#include <queue>#include <cstring>#include <algorithm>#include <set>using namespace std;int main(){int N,M,i;cin>>N;int distance[100002],temp;distance[1]=0;for(i=1;i<=N;i++){scanf("%d",&temp);distance[i+1]=distance[i]+temp;}cin>>M;int total=distance[N+1];while(M--){int x,y;scanf("%d%d",&x,&y);int minPos=min(x,y);int maxPos=max(x,y);int clockwise,anti;clockwise=distance[maxPos]-distance[minPos];anti=total-clockwise;int result=min(clockwise,anti);cout<<result<<endl;}}