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:

310
7
在读入每个出口与下个出口的距离的时候，也要对这些距离进行加和，从而可以减少计算两个特定的出口之间的最短距离的计算量。达到一劳永逸的效果。否则，当测试点很多的时候，运行时间可能达不到要求，最后一个测试点通不过。
#include <iostream>#include <vector>using namespace std;int main(){    vector<int> dis;    int num_exit;    cin >>num_exit;    dis.push_back(0);    int total_dis=0;    for(int i=1;i<num_exit+1;i++)    {        int tmp;        cin >>tmp;        total_dis+=tmp;        dis.push_back(total_dis);    }    int n;    cin >>n;    for(int i=0;i<n;i++)    {        int st,en;        cin >>st >>en;        int a,b;        a=max(st,en);        b=min(st,en);        int dis1,dis2;        dis1=dis.at(a-1)-dis.at(b-1);        dis2=dis.back()-dis1;        if(dis1<dis2)            cout <<dis1 <<endl;        else            cout <<dis2 <<endl;    }    return 0;}