pat 1046. Shortest Distance (20)

1046. Shortest Distance (20)

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 ... 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 (<=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

解：这一题我最初用的是Floyd，但是会卡内存，100000*100000的数组想都不敢想，然后就一直卡，后来想想20分的题目应该用不到Floyd算法，把问题简化了，我把路径想象成一条带子，拿测试样例举例，1-5，这条带子就是1-2-3-4-5-1-2-3-4，开一维数组，数组最大也不超过2*100000，根据起点和终点相减（这里我用的for循环计算，比如说4和1的最短路径，我就是for（i=1;i<4;i++），后来证明错误问题就出现在这里），得到绝对值temp，再得到边数之和sum-temp和temp中的较小值，即为目标最短距离值（这一点，和正确的思路已经相差不多了，好痛心，脑袋太笨），但是pat的第三个测试会超时。

后来百度了下别人的做法，http://www.2cto.com/kf/201311/254357.html，感觉很受用，我的问题就在于没有在输入的时候就存储从1开始到各点之间的距离值，导致后来还要用for循环计算temp。如果顶点数n值很大的话，似乎确实很耗时。

自己错误在：我用数组存储相邻边的距离，如1-2的距离，2-3的距离，3-4的距离等，而不是1-2，1-3，1-4的距离，导致后来需要使用for循环累加求解temp。

下面贴上代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;

int main()
{
    int n,m;  //n表示边数，m表示求最短路的测试个数
    while(cin>>n)
    {
        int *a=new int[n+3];  //动态生成
        int temp=0;   //临时存储边值，初始化
        int sum=0;    //所有边值之和，初始化
        a[1]=0;
        for(int i=1;i<=n;i++)
        {
            cin>>temp;
            sum+=temp;
            a[i+1]=a[i]+temp;
        }
        cin>>m;
        int beg,en;  //起点beg和终点en
        for(int i=1;i<=m;i++)
        {
            cin>>beg>>en;
            temp=abs(a[beg]-a[en]);
            cout<<min(temp,sum-temp)<<endl;
        }
    }
}