HDU 5422 Rikka with Graph】的更多相关文章

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,…
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…
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 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 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…
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…
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…