PAT A1046 Shortest Distance

缺乏、安全感 2021-10-01 05:54 179阅读 0赞

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。