PAT (Advanced Level) Practise 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

题意：给你一个环形路上相邻两点的距离，求任意两点之间的最短路

解题思路：求出距离的前缀和即可

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <stack>#include <queue>#include <vector>#include <map>using namespace std;#define ll long longint n,k;int sum[100090];int main(){    while(~scanf("%d",&n))    {        memset(sum,0,sizeof sum);        int x;        for(int i=2;i<=n+1;i++)        {            scanf("%d",&x);            sum[i]=sum[i-1]+x;        }        scanf("%d",&k);        int u,v;        while(k--)        {            scanf("%d %d",&u,&v);            if(u>v) swap(u,v);            printf("%d\n",min(sum[v]-sum[u],sum[u]+sum[n+1]-sum[v]));        }    }    return 0;}