PAT A1046 Shortest Distance

最新推荐文章于 2021-02-20 16:15:29 发布

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本文介绍了一个算法问题，即计算高速公路形成简单环形上的任意两个出口之间的最短距离。输入包括环形公路的出口数量和各段距离，以及若干对出口编号，任务是输出每对出口间的最短距离。解决方案采用预处理技术，避免了暴力算法的超时风险。

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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 [3,10
的五次方 ]), 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 (≤10的四次方 ), 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 10的7次方 .

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

AC代码如下：

#include <stdio.h>
const int MAXN = 100010;
//dis[i]表示1号节点按顺时针方向到达i号节点的顺时针方向的下一个节点的距离
//A[i]表示i号与i+1号顶点的距离 

int main(int argc, char *argv[]) {
	int dis[MAXN],A[MAXN];  
	int sum = 0,query,n,i,t,left,right,temp;
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		scanf("%d",&A[i]);
		sum+=A[i];   //累加sum
		dis[i]=sum;  //预处理dis数组 
	}
	scanf("%d",&query);
	for(i=0;i<query;i++){  //query个查询 
		scanf("%d%d",&left,&right);
		if(left>right){  //防止left>right 
			t=left;
			left=right;
			right=t;
		}  
		temp=dis[right-1]-dis[left-1];
		if(temp<(sum-temp)){
			printf("%d\n",temp);
		}
		else{
			printf("%d\n",sum-temp);
		}
	} 
	return 0;
}

注意：
1.不能使用暴力无脑算法，那样做会超时。
2.dis[i]表示1号节点按顺时针方向到达i号节点的顺时针方向的下一个节点的距离
3.A[i]表示i号与i+1号顶点的距离
4.dis数组和sum在读入的时候就可以累加得到，这样查询的时候其实就是dis[right-1]-[left-1],这样可以减少算法的时间复杂度。