Rikka with Graph  Accepts: 123  Submissions: 525  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others) 问题描述 众所周知,萌萌哒六花不擅长数学,所以勇太给了她一些数学问题做练习,其中有一道是这样的: 给出一张 nn 个点 n+1n+1 条边的无向图,你可以选择一些边(至少一条)删除. 现在勇太想知道有多少种方案使得删除之后图依然联通.…

Rikka with Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 182    Accepted Submission(s): 95 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation,…

Rikka with Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 190    Accepted Submission(s): 78 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation,…

Rikka with Graph 思路: 官方题解: 代码: #include<bits/stdc++.h> using namespace std; #define ll long long int main() { int t; scanf("%d",&t); while(t--) { ll n,m,ans; scanf("%lld%lld",&n,&m); ) { ans=m*+(m-)*m*+(n*(n-)-m*-(m-)…

Rikka with Graph 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5631 Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has a non-direct gra…

Rikka with Graph II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1051    Accepted Submission(s): 266 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situati…

Rikka with Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is…

Problem Description   As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has a non-direct graph with n vertices and n edges. Now he wants you to tell him…

Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has a non-direct graph with n vertices and m edges. The length of each edge . Now…

Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: For an undirected graph G with n nodes and m edges, we can define the distance between …

题目链接 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: For an undirected graph G with n nodes and m edges, we can define the distance bet…

Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has a non-direct graph with n vertices and m edges. The length of each edge is 1.…

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has a non-direct graph with n vertices and n+1 edges. Rikka can choose some of the edges (at least one…

http://acm.hdu.edu.cn/showproblem.php?pid=5424 哈密顿通路:联通的图,访问每个顶点的路径且只访问一次 n个点n条边 n个顶点有n - 1条边,最后一条边的连接情况: (1)自环(这里不需要考虑): (2)最后一条边将首和尾连接,这样每个点的度都为2: (3)最后一条边将首和除尾之外的点连接或将尾和出尾之外的点连接,这样相应的首或尾的度最小,度为1: (4)最后一条边将首和尾除外的两个点连接,这样就有两个点的度最小,度都为1 如果所给的图是联通的话,那…

题目大意: 在 N 个点 N 条边组成的图中判断是否存在汉密尔顿路径. 思路:忽略重边与自回路,先判断是否连通,否则输出"NO",DFS搜索是否存在汉密尔顿路径. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<queue> #include<algorithm> #include<cmath&g…

n个点最少要n-1条边才能连通,可以删除一条边,最多删除2条边,然后枚举删除的1条边或2条边,用并查集判断是否连通,时间复杂度为O(n^3) 这边犯了个错误, for(int i=0;i<N;i++){ fa[i]=i; } 这个将i<=N,导致错误,值得注意 AC代码: #pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio>…

如果原图不连通,直接输出0. 如果原图连通,删除X条边之后要保证新图连通,再看数据是n+1条边-->因此,最多只能删去两条边. 因为n=100,可以枚举进行验证,枚举删去每一条边是否连通,枚举删去每两条边是否连通,验证是否连通可以用并查集,可以BFS. #include<cstdio> #include<cstring> #include<cmath> #include<string> #include<vector> #include&l…

给你n个点,让你连m条边,使得任意两两点对之间的最短路的和最小(两点若不可达,最短路记作n). 初始时ans=n*n*(n-1). 先尽量连成菊花图,每连一次让答案减小2*((n-2)*(i-1)+(n-1)),i为当前菊花图中的点数. 连完后剩下的边,每连一次让答案减小2. 如果已经用了n*(n-1)/2条边,剩下的边没有意义. #include<cstdio> #include<algorithm> using namespace std; typedef long long…

题目传送门 题意:判断是否为哈密顿图 分析:首先一种情况是不合法的:也就是度数为1的点超过2个:合法的有:,那么从度数为1的点开始深搜,如果存在一种走法能够走完n个点那么存在哈密顿路 收获:学习资料 代码: /************************************************ * Author :Running_Time * Created Time :2015-8-29 20:37:34 * File Name :C.cpp *******************…

[Link]:http://acm.hdu.edu.cn/showproblem.php?pid=6090 [Description] 给你n个点; 让你在这n个点上最多连m条无向边; 使得 ∑ni=1∑nj=1dis(i,j)最小 这里dis(i,j)定义为点i到点j的最短距离; (距离定义为走过的边数目); [Solution] 如果还没有组成一棵树的话; 就尽量弄成菊花图; ↓ 这样加入一个点的时候,和所有的在这个连通块里面的点最远距离只为2; 然后,当边的个数大于n-1,也就是说,已经够…

题解:考虑贪心地一条一条边添加进去. 当 m \leq n-1m≤n−1 时,我们需要最小化距离为 nn 的点对数,所以肯定是连出一个大小为 m+1m+1 的联通块,剩下的点都是孤立点.在这个联通块中,为了最小化内部的距离和,肯定是连成一个菊花的形状,即一个点和剩下所有点直接相邻. 当 m > n-1m>n−1 时,肯定先用最开始 n-1n−1 条边连成一个菊花,这时任意两点之间距离的最大值是 22.因此剩下的每一条边唯一的作用就是将一对点的距离缩减为 11. 这样我们就能知道了最终图的形状了…

Rikka with Graph  Accepts: 123  Submissions: 525  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others) 问题描述 众所周知,萌萌哒六花不擅长数学,所以勇太给了她一些数学问题做练习,其中有一道是这样的: 给出一张 nn 个点 n+1n+1 条边的无向图,你可以选择一些边(至少一条)删除. 现在勇太想知道有多少种方案使得删除之后图依然联通.…

Rikka with Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 118    Accepted Submission(s): 52 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, s…

Rikka with Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2377    Accepted Submission(s): 766 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation…

1001 Rikka with Chess ans = n / 2 + m / 2 1002 Rikka with Graph 题意:n + 1条边,问减去至少一条使剩下的图连通的方案数. 分析:原来暴力选一条或两条就行了,脑子笨了.判连通用BFS或并查集,此题并查集更好 #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector…

HDU6095——Rikka with Competition 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6095 题目意思:抱歉虽然是签到题,现场真的没做出来,因为题目没看懂,题目说明现在给出n个选手的能力值,现在举办n-1场比赛,如果每场比赛中选出的两名选手之间能力的差大于k,则能力高的人获胜,否则两个人都获胜,问最后有多少人获胜了.(注:如果他如果有人可以击败他,则就不算获胜) 代码:从代码里面我们看出,先对n个选手的能力值排序,然后ct…

Rikka with Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 239    Accepted Submission(s): 157 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation,…

原文链接https://www.cnblogs.com/zhouzhendong/p/NowCoder-2018-Summer-Round10-F.html 题目传送门 - https://www.nowcoder.com/acm/contest/148/F 题意 给定一个完全图 $G$ ,有边权. 定义其线图的一条边的权值为“该边连接的两个点,在原图中对应的边 的权值和”. 在图 $L(G)$ 上,定义 $dis(i,j)$ 为节点 $i,j$ 之间的最短路. 求 $\sum_{i=1}^{n…

题意: 现在有一副完全图, 将他转化成线图. 线图就是 把原来的图上的边都变成点, 然后如果原来的任意2条边存在公共点, 他们就会有一条边, 边权为原来的2条边的和. 最后求出线图中的任意2点的最短路的和. 题解: 对于原图来说 现在有 ( a, b)  ->  (c, d)  他的花费是 W(a,b) + W(c, d) + 2*min( d[a][c], d[a][d], d[b][c], d[b][d] ) 的值.其中d[i][j] 代表 从i走到j的最小花费. 现在我们枚举起点(i, j…

Hello everyone! I am your old friend Rikka. Welcome to Xuzhou. This is the first problem, which is a problem about the minimum spanning tree (MST). I promise you all that this should be the easiest problemeasiest problem for most people. A minimum sp…