HDU 6343 Graph Theory Homework(水题)

Problem Description 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 ⌊|wi−wj|−−−−−−−√⌋ . 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 贴过来后公式混乱，题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6343 题目大意：给出边权求最短路。 思路：嗯，最短路，套模板，dij？不行，10^5，SPFA？来一发，，套模板也超时？？？看神仙的提交记录，0ms？看来有问题。然后发现，，，两点之间直线最短...(题解中说的是三角形两边和大于第三边)。 代码如下： #include<iostream> #include<algorithm> #include<cstdio> #include<cmath> using namespace std; int k[100005]; int main() { int n,m; scanf("%d",&n); while(n--){ scanf("%d",&m); for(int i=0;i<m;i++) scanf("%d",&k[i]); //sort(k,k+m); int b=abs(k[m-1]-k[0]); int c=int(sqrt(b)); printf("%d\n",c); } }