1046. Shortest Distance (20)-PAT甲级

题目：

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

这道题画个图会理解的很清楚，需要注意的是时间要求是100ms，所以说

一定要注意尽量不要让计算重复，怎么做那？

就是把它存起来，用空间换取时间。答案中就是用dis数组来储存距离的和

计算的时候只要一个减法就够了。

为了让计算统一，我们可以按顺时针来计算，要求最小值只要取得”,min(temp,sum-temp))就行了。

解答：

#include<cstdio>
#include<algorithm>
using namespace std;
const int MAXN=100005;
int dis[MAXN],A[MAXN];//A[i]表示i到i+1位置的距离
int main(){
int sum=0,n,left,right;
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",&A[i]);
sum+=A[i];
dis[i]=sum;//dis[i]预存1到i位置的距离之和，减少了以后的计算
}
int query;
scanf("%d",&query);
while(query--){
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));//一定有两条线路，取其中的最小距离
}
return 0;
}

2017/8/23添加

#include<cstdio>
#include<vector>
using namespace std;
int main(){
int n,m;
scanf("%d",&n);
vector<int> d(n+1);
for(int i=1;i<=n;i++){
scanf("%d",&d[i]);
d[i]+=d[i-1];
}
scanf("%d",&m);
while(m--){
int ans,a,b;
scanf("%d%d",&a,&b);
if(a>b)swap(a,b);
ans=d[b-1]-d[a-1];
printf("%d\n",min(d
-ans,ans));
}
return 0;
}