PAT A1046. Shortest Distance (20)
PAT A1046. Shortest Distance (20)
阿豪boy 发表于10个月前
PAT A1046. Shortest Distance (20)
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腾讯云 技术升级10大核心产品年终让利>>>   

https://www.patest.cn/contests/pat-a-practise/1046

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 ... 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.

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 9
3
1 3
2 5
4 1

Sample Output:

3
10
7

 

#include <iostream>
#include <cstdio> 

//dis[i] 表示 由 1出发到i的距离 
//由a到b的距离就是 dis[b]-dis[a] 默认a<b 
  
int dis[100010];
using namespace std;
int main(int argc, char *argv[])
{
	int n,t,m;
	scanf("%d",&n);
	dis[1]=0; 
	int sum=0;  //环的总长 
	for(int i=2;i<=n+1;i++){
		scanf("%d",&t);
		dis[i]=dis[i-1]+t;	
		sum += t;
	}
	scanf("%d",&m);
	while(m--){
		int a,b;
		scanf("%d%d",&a,&b);
		if(a>b) swap(a,b);
		printf("%d\n",min( dis[b]-dis[a],sum-(dis[b]-dis[a])) ); 	
	} 
	return 0;
}

 

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