题目链接:https://vjudge.net/problem/HDU-4081 Qin Shi Huang's National Road System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8970    Accepted Submission(s): 3175 Problem Description During the…

先发发牢骚:图论500题上说这题是最小生成树+DFS,网上搜题解也有人这么做.但是其实就是次小生成树.次小生成树完全当模版题.其中有一个小细节没注意,导致我几个小时一直在找错.有了模版要会用模版,然后慢慢融会贯通.我要走的路还长着啊. 这里有两个次小生成树的模版: http://www.cnblogs.com/Potato-lover/p/3949996.html 此题的解题思想:在prim算法中做一些改变,求出任意两点(u,v)路径之间的最大权值,并记录,记为maxe[u][v].运行一遍pr…

题目:Qin Shi Huang's National Road System Qin Shi Huang's National Road System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2049    Accepted Submission(s): 746 Problem Description During the W…

Qin Shi Huang's National Road System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) [Problem Description] During the Warring States Period of ancient China(476 BC to 221 BC), there were seven kingdoms in China ---…

题意: 秦始皇要建路,一共有n个城市,建n-1条路连接. 给了n个城市的坐标和每个城市的人数. 然后建n-2条正常路和n-1条魔法路,最后求A/B的最大值. A代表所建的魔法路的连接的城市的市民的人数的和,B 代表n-2条正常路的长度的和. 思路: 这题是次小生成树的变形,所谓次小生成树的核心应该是记录树上节点间的路中最大的边的权重,然后将这条边替换为非最小生成树中的边,然后枚举找到最小值. #include<stdio.h> #include<string.h> #include…

Qin Shi Huang's National Road System                                                                   Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)                                                                …

Qin Shi Huang's National Road System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10567    Accepted Submission(s): 3727 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4081 Description: During…

题目链接: Qin Shi Huang's National Road System Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 32768/32768 K (Java/Others) Problem Description   During the Warring States Period of ancient China(476 BC to 221 BC), there were seven kingdoms in Ch…

枚举作为magic road的边,然后求出A/B. A/B得在大概O(1)的时间复杂度求出,关键是B,B是包含magic road的最小生成树. 这么求得: 先在原图求MST,边总和记为s,顺便求出MST上任意两点路径上的最长边d[i][j]. 当(u,v)是magic road时, 如果它在原本的MST上,则B就等于s-原(u,v)的权,而原(u,v)的权其实就是d[u][v]: 如果它不在原本的MST上,则B就等于s-d[u][v]+0. 总之就是一个式子:B=s-d[u][v]. 于是,在…

先求最小生成树 再遍历每一对顶点,如果该顶点之间的边属于最小生成树,则剪掉这对顶点在最小生成树里的最长路径 否则直接剪掉连接这对顶点的边~ 用prim算法求最小生成树最长路径的模板~ #include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include<cmath> using namespace std; ; const int inf=1e9; dou…