PAT A1046 Shortest Distance (20)
来源:互联网 发布:java写倒三角 编辑:程序博客网 时间:2024/05/16 11:35
题目地址:https://www.patest.cn/contests/pat-a-practise/1046
题目描述:
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.
输入格式(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.
输出格式(Output Specification):
For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.
输入样例(Sample Input):
5 1 2 4 14 9
3
1 3
2 5
4 1
输出样例(Sample Output):
3
10
7
题意:
有N个结点围城一个圈,相邻两个点之间的距离已知,且每次只能移动到相邻点。然后给出M个询问,每个询问给出两个数字A和B即结点编号(1<=A,B<=N),求从A号结点到B号结点的最短距离。
样例解释:
如图所示,共有5个结点,分别标号为1、2、3、4,相邻两点的距离在图上给出。总共三个询问:
13:从1号点到3号点的最短距离为3,路径为1->2->3;
25:从2号点到5号点的最短距离为10,路径为2->1->5;
41:从4号点到1号点的最短距离为7,路径为4->3->2->1。
解题思路:
步骤 1:以dis[ i ]表示 1 号结点按顺时针方向到达”i 号结点顺时针方向的下一个结点“的距离(1 <= i <= N),sum 表示一圈的总距离。于是对每个查询 left ->right,其结果就是 dis( left , right ) 与 sum - dis(left , right)中的较小值。
步骤 2:dis 数组和 sum 在读入时就可以进行累加得到。这样对每个查询 left -> right,dis(left,right)其实就是 dis[ right - 1] - dis[left - 1]。这样可以做到查询复杂度为O(1)。
注意:
- 查询的两个点的编号可能会有 left > right 的情况。这样情况下,需要交换 left 和 right
- 此题若没有经过预处理 dis 数组和 sum 的做法会很容易超时
- 之所以不把 dis[ i ] 设置为 1 号结点按顺时针方向到达 i 号结点的距离,是因为 N 号结点到达 1 号结点的距离无法被这个数组所保存
C++完整代码如下:
#include<cstdio>#include<algorithm>using namespace std;const int MAXN = 100005;int dis[MAXN], A[MAXN]; // dis数组含义已说明,A[I] 存放 i 号与 i+1 号顶点的距离int main(){ int sum = 0, query, n, left, right; scanf("%d", &n); for(int 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(int i = 0; i < query; i++){ //query 个查询 scanf("%d%d", &left, &right); // left -> right if(left > right) swap(left, right); //left -> right 时交换 int temp = dis[right - 1] - dis[left - 1]; printf("%d\n", min(temp, sum - temp)); } return 0;}
- PAT A1046 Shortest Distance (20)
- 【PAT】A1046. Shortest Distance (20)
- PAT A1046. Shortest Distance (20)
- PAT A1046. Shortest Distance
- A1046. Shortest Distance (20)
- PAT-A1046 Shortest Distance(模拟)
- 3.1简单模拟A1046. Shortest Distance (20)
- A1046.Shortest Distance
- PAT A1046 Shortest Distances
- 【PAT】1046.Shortest Distance (20)
- 【PAT】1046. Shortest Distance (20)
- PAT 1046. Shortest Distance (20)
- PAT A Shortest Distance (20)
- PAT 1046. Shortest Distance (20)
- PAT 1046. Shortest Distance (20)
- pat 1046. Shortest Distance (20)
- PAT 1046. Shortest Distance (20)
- 【PAT】1046. Shortest Distance (20)
- CTF Field Guide
- Spring的注解驱动控制程序
- 在mysql中关于通过多字段去除重复记录的方法
- AIDL
- androidapp 检测到移动物体,通过web在电脑上实时显示出来
- PAT A1046 Shortest Distance (20)
- web每章重点
- MySQL数据库学习笔记(三)----基本的SQL语句
- Java学习笔记之反射(Reflection)
- 萌新的Canvas笔记(一)
- 【ASP.NET】——Web服务器控件、HTML服务器控件
- hihocoder 1139 二分答案+BFS求最短路
- 理财先理观念
- 多线程生成海量数据和多线程海量数据查找