PAT (Advanced Level) Practice 1046 Shortest Distance (20 分) 凌宸1642

PAT (Advanced Level) Practice 1046 Shortest Distance (20 分) 凌宸1642

题目描述:

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.

译:你的任务很简单:给定 N 个出口,形成一个简单的圆形公路,你应该说出任意一对出口之间的最短距离。


Input Specification (输入说明):

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 ⋯ DN, where Di is the distance between the i-th and the ( i +1 )-st exits, and DN 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.

译:每个输入文件包含一个测试用例,每个用例在第一行中包含一个正整数 N ( 3 ≤ N ≤10 5 ) , 紧跟着 N 个表示距离的整数 D1 D2 ⋯ DN , Di 表示 第 i 个 出口到第 i + 1 个出口之间的距离, DN 表示第 N 个出口到第 1 个出口之间的距离。所有的数字被一个空格分隔。第二行给出一个正整数 M (≤104) , 接下来 M 行,每行包含一对出口的编号,保证出口在 1 ,N之间。题目保证整个环道的距离不超过 107


Output Specification (输出说明):

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

译:对于每个测试用例,在 M 行中打印相应那对出口之间的最短距离 。


Sample Input (样例输入):

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

Sample Output (样例输出):

3
10
7

The Idea:

本题的最短距离还算简单。我们只需要一个 数组存储 第 1 个出口 到 第 i 个出口之间的距离。然后求两个出口之间的距离,就变成了简单的减法问题,由于是一个环道,最短距离需要考虑在两个距离之间抉择:a 到 b 的距离 和整个环道的距离 减去 a 到 b 的距离 。


The Codes:

#include<bits/stdc++.h>
using namespace std ;
#define MAX 100010
int sum[MAX] = { 0 } ; 
int n , m , t , a , b ;
int main(){
	scanf("%d" , &n) ;
	for(int i = 1 ; i <= n ; i ++){
		scanf("%d" , &t) ;
		sum[i] = sum[i-1] + t ; // 计算 第 1 个出口 到 第 i 个出口之间的距离。
	}
	scanf("%d" , &m) ;
	while(m --){
		scanf("%d%d" , &a , &b) ;
		if(a > b) swap(a , b) ; // 如果 a 大于 b 就交换一下
		cout<<min(sum[b - 1] - sum[a - 1] , sum[n] - (sum[b - 1] - sum[a - 1]))<<endl ;
	}
	return 0;
}

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