PAT A1046超时

最新推荐文章于 2022-07-14 22:58:08 发布

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最新推荐文章于 2022-07-14 22:58:08 发布

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该博客讲述了PAT A1046题目的解决方案，涉及寻找给定高速公路出口间的最短距离。初始思路导致运行超时，优化方案是存储从节点1到各节点的顺时针距离，以O(1)时间复杂度计算任意两点间的最短距离，避免了超时问题。

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1046 Shortest Distance （20 分）

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

一开始思考的思路比较简单，无非就是用数组a保存相邻节点间的距离，然后对于m组x，y的结点的距离查询，先进行x>y情况下的交换，使x<y，之后累加x与y之间的结点距离即得到顺时针下的x-->y的距离d1，则sum-d1即为逆时针下的y-->x的距离d2,之后输出更小的一个。代码如下：

#include <stdio.h>

int main(){
    int n;
    scanf("%d",&n);
    int a[n];     //a[i]表示i+1-->i+2的距离
    int sum=0;    //组成的圆的距离和
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
        sum+=a[i];
    }
    int m;
    scanf("%d",&m);
    int x,y;
    for(int i=0;i<m;i++){
        scanf("%d%d",&x,&y);
        if(x==y)
            printf("0");
        else{
            if(x>y){
                int temp=x;x=y;y=temp;
            }
            int d1=0;
            for(int i=x-1;i<y-1;i++)
                d1+=a[i];     
            if(d1<sum-d1)
                printf("%d",d1);
            else
                printf("%d",sum-d1);
        }
        if(i!=m-1)
            printf("\n");
    }
}

但结果是运行超时，问题在于每一组x,y都要进行x与y之间所有节点间距离的累加，因此不再使用数组a保存相邻两节点的距离，而是使用dis[]保存结点1至各节点的顺时针距离（dis[i]表示1-->i+1结点的距离），这样计算d1=dis[y-1]-dis[x-1]，时间复杂度为O(1)，不会出现超时的情况。代码如下：

#include <stdio.h>
int main(){
    int n;
    scanf("%d",&n);
    int dis[n+1]; //dis[i]表示1-->i+1结点的距离
    dis[0]=0; //1-->1=0
    int a;
    int sum=0;
    for(int i=0;i<n;i++){
        scanf("%d",&a);
        dis[i+1]=dis[i]+a;
        sum+=a;
    }
    int m;
    scanf("%d",&m);
    int x,y;
    for(int i=0;i<m;i++){
        scanf("%d%d",&x,&y);
        if(x>y){
                int temp=x;x=y;y=temp;
            }
        int d1=dis[y-1]-dis[x-1];
        if(d1<sum-d1)
            printf("%d",d1);
        else
            printf("%d",sum-d1);
        if(i!=m-1)
            printf("\n");
    }
}