PAT A1046 1046 Shortest Distance (20 分)

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 D₁D₂⋯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 9 3 1 3 2 5 4 1

Sample Output:

3 10 7

题意：

求点到点之间的最短路径；

输入格式：

节点数N，两节点之间的距离D[i]（i到i+1之间的距离） 测试数目 节点1 节点2 节点1 节点2 … …

输出格式：

测试点1的最短距离 测试点2的最短距离 …

思路：

（1）开一个数组dis,dis[i]存放节点1到节点i+1的距离，sum表示所有节点之间的路径总和； （2）节点left到节点right的最短距离为dis[right-1]-dis[left-1]与sum-(dis[right-1]-dis[left-1])之间的最小值；

代码：

#include <cstdio> #include <algorithm> using namespace std; int main(){ int N; scanf("%d",&N); int D[N+1],sum=0,dis[N+1]; dis[0]={0}; for(int i=1;i<=N;i++){ scanf("%d",&D[i]); sum+=D[i]; dis[i]=sum;//dis[i]存放1到i+1的距离 } int M; scanf("%d",&M); int left,right,dis_result; for(int i=0;i<M;i++){ scanf("%d %d",&left,&right); if(left>right){ int temp=left; left=right; right=temp; } dis_result=dis[right-1]-dis[left-1]; dis_result=min(dis_result,sum-dis_result); printf("%d\n",dis_result); dis_result=0; } return 0; }

词汇：

a positive integer 一个正整数 corresponding 相应的