A1046. Shortest Distance

题目描述#

　　The task is really simple: given N exits on a high way which forms a simple cycle, you are supposed to tell the shortest distance between any pair of exits

输入格式#

　　Each input file contains on 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 D₁ 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 exists  are numbered from 1 to N. It is guaranteed that the total round trip distance is no more than 10⁷

输出格式#

　　For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

输入样例#

　　5 1 2 4 14 9

　　3

 　   1 3

　　2 5 

　　4 1

输出样例#

　　3

　　10 

　　7 

题意#

　　N个结点围成一个圈，相邻两个点之间的距离已知，且每次只能移动到相邻点。然后给出M个询问，每个询问给出两个数字A和B，即结点编号(1≤A，B≤N)，求从A

号结点到B号结点最短距离

输入格式#

　　每个输入文件包含一个测试用例

对于每种情况，第一行都包含一个整数N（在[3，10⁵]中），后跟n个整数距离D₁ D₂ ... D_N，其中D₁是第i个与（i + 1）之间的距离出口，而D_N在第N个出口和第一个出口之间。

一行中的所有数字都用空格分隔。 第二行给出一个正整数M（≤10⁴），紧随其后的M行，每行包含一对出口号，条件是出口的编号从1到N。确保本地往返距离不超过 10⁷

 

输入样例，有3个查询，分别是1，3；2，5；4，1

解题思路#

• 这是一个圆，圆上两端点间距离的最小值，就是两点间的劣弧

const int MAXN = 100000;
int dis[MAXN], A[MAXN];// dis 数组含义已说明，A[i] 存放i号与i+1 号顶点的距离
int main(int argc, char *argv[]) {
    int sum = 0, query, n, left, right;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
        scanf("%d", &A[i]);
        sum += A[i];
        dis[i] = sum;// 初始化dis数组 
    } 
    scanf("%d", &query);
    for(int i = 0; i < query; i++){
        scanf("%d%d", &left, &right);
        if(left > right){
            swap(left, right);
        }
        int temp = dis[right - 1] - dis[left - 1];// 两端点沿顺时针的方向的弧 
        printf("%d\n", min(temp, sum - temp));// 选择劣弧输出 
    }
    return 0;    
}