PAT

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

分析：

题目：城市之间高速依次连成一个圈，给出相邻距离，求给定城市的最短距离。解题：输入时，记录距离和，在求两个城市的距离是，对记录的sumi sumj做差，求余。有两种情况，由于是个圈必定左边一条，右边一条，要比较所求距离结果，输出最短。

code:

#include<iostream> #include<cstdio> #include<vector> using namespace std; vector<int>dis; int main() { freopen("in","r",stdin); int N,M,tmp,tmp2,sum; scanf("%d",&N); sum=0; dis.push_back(sum);//city0 起始距离0 for(int i=0;i<N;i++) { scanf("%d",&tmp); sum+=tmp; dis.push_back(sum);//记录距离和，求两个城市距离通过做差求得 } scanf("%d",&M); int a,b; for(int i=0;i<M;i++) { scanf("%d%d",&tmp,&tmp2); tmp--;tmp2--;//纠正数组下标 a=(dis[tmp]-dis[tmp2]+sum)%sum;//避免出现负数，求补 b=(dis[tmp2]-dis[tmp]+sum)%sum; if(a<b) printf("%d\n",a); else printf("%d\n",b); } return 0; } AC: