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.
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
分析
测试点三超时思路：刚开始想的是将编数据存放到数组中，下标就作为顶点，查找的过程中就可以直接累加start-end的数组数据就可以了，中途也是要判断start和end数字的大小，还有就是输出的是哪一个圈圈，刚开始是手写比大小和交换数字大小，后来才明白其实是可以用swap和min函数的。超时原因就是外面一层循环控制输入数据的组数，内层还有一个遍历数组的for循环。
改进
改进就是时用一个叫做dis的数组进行存储第一个顶点到第i+1个顶点的距离，这个具体描述参考算法详解具体介绍，我觉得这个的好处就是，可以省略掉遍历数组的操作，dis数组中直接存储的是距离，之后直接使用dis[right-1]-dis[left-1]就可以了，通俗一点的理解就是可以将她想象成一段线段，要求的就是中间的一段长度，就是用当前位置的总长度-截取线段长度的开始数值就可以了。
超时代码
#include<bits/stdc++.h>
using namespace std;
int main()
{
ios::sync_with_stdio(false);cin.tie(0);int a[100000];int n,m,start,end;int sum=0,ans1=0,asn2=0;scanf("%d",&n);for(int i=1;i<=n;i++){
scanf("%d",&a[i]);sum+=a[i];}scanf("%d",&m);for(int i=0;i<m;i++){
ans1=0;scanf("%d %d",&start,&end);if(start>end){
int temp;temp=start;start=end;end=temp;}for(int j=start;j<end;j++){
ans1+=a[j];}if(ans1<sum-ans1){
printf("%d",ans1);}else {
printf("%d",sum-ans1);}if(i<m-1){
printf("\n");}}return 0;
}
正确代码
#include<bits/stdc++.h>
using namespace std;const int MAXN=100005;
int dis[MAXN],A[MAXN]={
0};int main(){
ios::sync_with_stdio(false);cin.tie(0);int sum=0;int query,n,left,right;scanf("%d",&n);for(int i=1;i<=n;i++){
scanf("%d",&A[i]);sum+=A[i];dis[i]=sum;}scanf("%d",&query);for(int i=0;i<query;i++){
scanf("%d %d",&left,&right);if(left>right){
swap(left,right);}int temp=dis[right-1]-dis[left-1];printf("%d\n",min(temp,sum-temp));}return 0;
}
总结
求最小值和交换数值时可以直接使用swap函数和min函数的，还有就是一种新的方法，求一段长度的时候，如果超时，就可以使用一个新的数组，记录每一个小终点到初始点的距离，这样子就可以解决超时问题。
