Shortest Path 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5636 Description There is a path graph G=(V,E) with n vertices. Vertices are numbered from 1 to n and there is an edge with unit length between i and i+1 (1≤i<n). To make the graph more in…

Shortest Path Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1146 Accepted Submission(s): 358 Problem Description There is a path graph G=(V,E) with n vertices. Vertices are numbered from 1 to…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5636 题解: 1.暴力枚举: #include<cmath> #include<cstdio> #include<iostream> #include<algorithm> using namespace std; typedef long long LL; ; ; int n, m; ], b[]; int main() { int T; scanf(&qu…

题意:bc round 74 分析(官方题解): 你可以选择分类讨论, 但是估计可能会写漏一些地方. 只要抽出新增边的端点作为关键点, 建立一个新图, 然后跑一遍floyd就好了. 复杂度大概O(6^2m) 注:然后我不会这种,这种floyd我觉得复杂度应该是复杂度应该是O(8^3m) 大概在千万级别,其实应该可以过,然后,其实只需要求单元最短路就行,然后是不是可以dij,然后就快一点 反正我也没写 我在比赛的时候写的是分治,考虑走不走新加的边每次走几条,以及走的顺序就好 然后全排列,时间复杂度…

题目链接  HDU5636 n个点,其中编号相邻的两个点之间都有一条长度为1的边,然后除此之外还有3条长度为1的边. m个询问,每次询问求两个点之前的最短路. 我们把这三条边的6个点两两算最短路, 然后询问的时候用这6个点的距离来更新答案就可以了. (不过听说好像有更好的方法,先占个坑) 时间复杂度$O(216m)$ #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <=…

题目链接:pid=3631" style="font-size:18px">http://acm.hdu.edu.cn/showproblem.php?pid=3631 Shortest Path Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3962    Accepted Submission(s): 9…

Shortest Path Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u SubmitStatus Description When YY was a boy and LMY was a girl, they trained for NOI (National Olympiad in Informatics) in GD team. One day, GD team's coach, Prof. G…

The Shortest Path in Nya Graph Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description This is a very easy problem, your task is just calculate el camino mas corto en un grafico, and just solo hay que camb…

floyd算法好像很奇妙的样子.可以做到每次加入一个点再以这个点为中间点去更新最短路,效率是n*n. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<algorithm> using namespace std; ; const int INF = 0x7FFFFFFF; int A[maxn][maxn], flag[maxn]; int…

题意:给定一个图,求从1到N的递增边权的最短路. 解法:类似于bellman-ford思想,将所有的边先按照权值排一个序,然后依次将边加入进去更新,每条边只更新一次,为了保证得到的路径是边权递增的,每次将相同权值的边全部取出来一同更新,每条边能够更新的前提是某一个端点在之前被更小的边权更新过.另外一个要注意的地方就是一次相同边的更新中,要把所有的更新暂存起来最后一起去更新,这样是为了防止同一权值的边被多次加入到路径中去. #include <iostream> #include <cst…