There is a complete graph containing n vertices, the weight of the i-th vertex is wi.
The length of edge between vertex i and j (i≠j) is .
Calculate the length of the shortest path from 1 to n
.
Input
The first line of the input contains an integer T
(1≤T≤10)
denoting the number of test cases.
Each test case starts with an integer n (1≤n≤105) denoting the number of vertices in the graph.
The second line contains n integers, the i-th integer denotes wi (1≤wi≤105)
.
Output
For each test case, print an integer denoting the length of the shortest path from 1
to n
.
Sample Input
1
3
1 3 5
Sample Output
2
从前一个点可以任意选择下一个点, 每个点有一个权值, 点和点之间的距离是,权值差的绝对值开根号再向下取整,求从1 到n最短距离。
刚开始做这个题, 觉得是个水题, (实际上确实是个水题),写了个 n ^ 2的记忆化搜索, 抱着侥幸心理交了一次, 不出意外的超时了~
实际上我们可以假设有某中间点k, 使得直接从1到n的路线松弛, 那么有
两边平方就会发现, k的存在是矛盾的,所以知道最短距离是从1直接到n。
AC代码:
#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int i, j, T, ans, n, q, m;
int s[100005];
scanf("%d", &T);
while(T--){
scanf("%d", &n);
for(i = 1;i <= n;i++)
scanf("%d", &a);
m = abs(s[1] - s[n]);
ans = (int)sqrt(m * 1.0);
printf("%d\n", ans);
}
return 0;
}