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 ... D_N, where D_i is the distance between the i-th and the (i+1)-st exits, and D_N 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 14 931 32 54 1

Sample Output:

3107

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分析：

（1）给定一个长为s环求环中两点的最短距离，其实就是一段距离d和s-d的小值。

（2）求d如果用循环，会超时。所以要在第一次输入循环时候，直接计算出一点到初始点的距离，之后求某两点距离直接相减可得。O(N2）不靠谱啊啊。

#include <stdio.h>int main (){int n,m,i ,circleDistance=0;int dt[100002],dt2[100002],op[10002];scanf("%d",&n);int temp1=0;dt2[0]=0;for (i=0;i<n;i++){scanf("%d",&dt[i]);dt2[i+1]=temp1+dt[i];temp1=dt2[i+1];circleDistance+=dt[i];}    scanf ("%d",&m);for (i=0;i<m;i++){int a,b;scanf ("%d%d",&a,&b);if (a>b){int temp=a;a=b;b=temp;}int sd1,sd2;sd1=dt2[b-1]-dt2[a-1];sd2=circleDistance-sd1;op[i]=sd1<sd2?sd1:sd2;}for (i=0;i<m;i++){printf ("%d\n",op[i]);}return 0;}