PAT 甲级 1046 Shortest Distance (20 分)(前缀和,想了一会儿)
作者:互联网
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]), followed by N integer distances D1 D2 ⋯ DN, where Diis the distance between the i-th and the (-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 (≤), 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 1.
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
题意:
给出一个环形的高速公路,其中有N个出口,第Di个出口是i到i-1的距离,而DN是N到1 的距离。给出任意两个出口,计算两者的最短距离。
题解:
显而易见,这是一个循环队列,计算距离需要考虑两个方向,但是如果直接遍历的话,时间复杂度为O(n2) O(n^2)O(n 2)这个数据量会超时(第三个测试点),所以我们需要考虑,如何优化这个距离计算过程。不妨考虑,计算每个出口两个方向的累加距离,这样计算两者之间的距离的时候,直接做加减即可,时间复杂度为O(n) O(n)O(n)。
AC代码:
#include<iostream> #include<algorithm> #include<vector> #include<queue> #include<map> #include<string> #include<cstring> using namespace std; int n; int a[100005]; int s1[100005]; int s2[100005]; int main() { cin>>n; memset(s1,0,sizeof(s1)); memset(s1,0,sizeof(s2)); for(int i=1;i<=n;i++){ cin>>a[i]; s1[i+1]=s1[i]+a[i]; } for(int i=1;i<=n;i++){ s2[i+1]=s2[i]+a[n-i+1]; } /*for(int i=1;i<=n;i++){ cout<<i<<" s1 "<<s1[i]<<endl; cout<<i<<" s2 "<<s2[i]<<endl; }*/ int m; int u,v; cin>>m; for(int i=1;i<=m;i++){ cin>>u>>v; if(u>v){ int temp=v; v=u; u=temp; } cout<<min(s1[v]-s1[u],s2[n+2-v]+s1[u])<<endl;//两个方向选最大 } return 0; }
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]), followed by N integer distances D1 D2 ⋯ DN, where Diis the distance between the i-th and the (-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 (≤), 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 1.
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
标签:Distance,contains,PAT,1046,exits,distance,int,between,each 来源: https://www.cnblogs.com/caiyishuai/p/11456499.html