PAT——甲级1046S：shortest Distance

这道题，折磨了我一个多小时，前前后后写了三个算法。

1046 Shortest Distance （20 point(s)）

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​5​​]), followed by N integer distances D​1​​D​2​​ ⋯ D​N​​, where D​i​​ is the distance between the i-th and the (i+1)-st exits, and D​N​​ 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​4​​), 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​7​​.

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

 题目特很简单，就是一个多个点组成的环形，求两个点之间的最短距离。阳历第一行5代表5个点，后面第i个数代表第一个点到第i+1个点的距离。等等。。。

然后我第一个想法是循环链表，第一次完成的写链表的代码，废了九牛二虎之力，写了一个循环链表，然后就贼舒服的求最短距离。

先贴出来的我写的代码

#include<cstdio>
class Node
{
public:
    int num;
    int dis;
    Node* next;
    Node* pre;
};

class cycleLinkList
{
public:
    cycleLinkList();
    void insert_back(int num, int dis);
    int find_nearest(int prenum,int lastnum);
    //void displayAll(void);
private:
    Node* head;
};
cycleLinkList::cycleLinkList() 
{ 
    head = new Node;
    head->num = 0;
    head->pre = head;
    head->next = 
    head;head->dis = 0; 
}
void cycleLinkList::insert_back(int num, int dis) {
    Node* pre = head->next;
    while (pre->next!=head)
    {
        pre = pre->next;
    }
    Node* new_Node=new Node;
    new_Node->dis = dis;
    new_Node->num = num;
    new_Node->next = head;
    new_Node->pre = pre;
    pre->next->pre = new_Node;
    pre->next = new_Node;
}
int cycleLinkList::find_nearest(int prenum, int lastnum) {
    Node* pre = head->next;
    while (pre->num != prenum)
    {
        pre = pre->next;
    }
    int predistance=0,nextdistance=0;
    Node* pre_dir = pre;
    while (pre_dir->num != lastnum)
    {
        pre_dir = pre_dir->pre;
        predistance += pre_dir->dis;
        
    }
    Node* next_dir = pre;
    while (next_dir->num != lastnum)
    {
        nextdistance += next_dir->dis;
        next_dir = next_dir->next;
    }
    if (nextdistance >= predistance)return predistance;
    else return nextdistance;

}
/*
void cycleLinkList::displayAll(void)
{
    Node* _head = head->next;
    while (_head != head)
    {
        printf("%d  %d\n", _head->num, _head->dis);
        _head = _head->next;
    }
    printf("pre\n");
    _head = head->pre;
    while (_head != head)
    {
        printf("%d  %d\n", _head->num, _head->dis);
        _head = _head->pre;
    }
}*/

int main()
{
    int n,m,dis;
    scanf("%d", &n);
    cycleLinkList data;
    for (int i = 0; i < n; i++){
        scanf("%d", &dis);
        data.insert_back(i + 1, dis);
    }
    //data.displayAll();
    //printf("insert");
    scanf("%d", &m);
    for (int i = 0; i < m; i++){
        int pre_num, last_num;
        scanf("%d%d", &pre_num,&last_num);
        printf("%d\n",data.find_nearest(pre_num, last_num));    
    }
    return 0;
}

写完之后运行很正常，当然是很舒服啦，但是。。。提交给OJ就是运行超时，我也很无奈，循环链表那么直观，竟然不让用。

那我就用数组呗，第二个想法，用数组。

 1 #include<cstdio>
 2 int main()
 3 {
 4     int dis[100005],n,m,total_dis=0;
 5     memset(dis, 0, (100005*sizeof(int)));
 6     scanf("%d", &n);
 7     for (int i = 0; i < n; i++)
 8     {
 9         scanf("%d", &dis[i]);
10         total_dis += dis[i];
11     }
12     scanf("%d", &m);
13     for (int i = 0; i < m; i++) {
14         int prenum, lastnum,nextdis=0;
15         scanf("%d%d", &prenum, &lastnum);
16         if (prenum > lastnum) { int temp = prenum;prenum = lastnum;lastnum = temp; }
17         
18         for (int j = prenum - 1;j < lastnum - 1;j++)
19             nextdis += dis[j];
20         if (total_dis - nextdis >= nextdis) printf("%d\n", nextdis);
21         else printf("%d\n", total_dis-nextdis);
22     }
23     return 0;
24 }

不过这个代码交上去，还是有一个运行超时。牛客PATOJ的全部通过，但PAT官网OJ最后一个超时。

主要原因出在我每次计算最短距离都要遍历数组，大概相当于O（m*n）的复杂度。运行时间太长。

然后我就求助教材了，发现可以给数组保存为第i个元素是第i个点到第i个点的距离，然后求两个点a,b之间的距离就可以直接dis[b]-dis[a]了。（其中a<b，否则交换）

这样的话就输入的时候遍历n次，找最短距离的时候遍历m次就行了。

 1 #include<cstdio>
 2 #include<algorithm>
 3 using namespace std;
 4 int main()
 5 {
 6     int dis[100005], n, m, total_dis = 0;
 7     scanf("%d", &n);
 8     for (int i = 0; i < n; i++)
 9     {
10         int temp;
11         scanf("%d", &temp);
12         dis[i] = total_dis;
13         total_dis += temp;
14         //printf("i: %d dis:%d  total:%d\n", i, dis[i],total_dis);
15     }
16     scanf("%d", &m);
17     for (int i = 0; i < m; i++) {
18         int prenum, lastnum;
19         scanf("%d%d", &prenum, &lastnum);
20         if (prenum > lastnum) swap(prenum, lastnum);
21         int temp = dis[lastnum-1] - dis[prenum-1];
22         printf("%d\n",min(temp,total_dis-temp));
23     }
24     return 0;
25 }

然后终于通过啦。

这也学到了一点，拿到题先想想有没有简化的方法，按照最简单的思维去写代码，往往可能运行超时。

就像这道题最后一种方法，就简化了很多。越简单的题越需要脑子，不要一拿上来就做。

int main(){int dis[100005],n,m,total_dis=0;memset(dis, 0, (100005*sizeof(int)));scanf("%d", &n);for (int i = 0; i < n; i++){scanf("%d", &dis[i]);total_dis += dis[i];}scanf("%d", &m);for (int i = 0; i < m; i++) {int prenum, lastnum,nextdis=0;scanf("%d%d", &prenum, &lastnum);if (prenum > lastnum) { int temp = prenum;prenum = lastnum;lastnum = temp; }for (int j = prenum - 1;j < lastnum - 1;j++)nextdis += dis[j];if (total_dis - nextdis >= nextdis) printf("%d\n", nextdis);else printf("%d\n", total_dis-nextdis);}return 0;}