1046. Shortest Distance (20)
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
题意：求一个环形路线的两两节点间最小距离。
思路：刚开始是最原始的一一累加，第三个case超时；后来想用DP，但在定义数组int dis[100001][100001]时报“数组太大”的错误；放弃DP了开始琢磨数据的规律，发现①环形的总长度是可求的②A到B之间的距离等于（A到0距离）减去（B到0距离）③任意节点到0的距离都可以在输入时求出。
代码如下：
#include <iostream>
#include <fstream>
using namespace std;
ifstream fin("in.txt");
#define cin fin
int main()
{
int n;
cin>>n;
int* dis = new int[n+1];
int d;
int i;
int sum = 0;
for(i=0;i<n;i++)
{
cin>>d;
dis[i+1] = sum;
sum = sum + d;
}
int m;
cin>>m;
int begin,end;
int total;
for(i=0;i<m;i++)
{
cin>>begin>>end;
if(begin>end)
{
total = dis[begin]-dis[end];
}else
{
total = dis[end]-dis[begin];
}
if(sum-total > total)
{
cout<<total<<endl;
}else
{
cout<<sum-total<<endl;
}
}
system("PAUSE");
return 0;
}