HDU 5424——Rikka with Graph II——————【哈密顿路径】
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
Yuta has a non-direct graph with n vertices and n edges. Now he wants you to tell him if there exist a Hamiltonian path.
It is too difficult for Rikka. Can you help her?
For each testcase, the first line contains a number n(1≤n≤1000).
Then n lines follow. Each line contains two numbers u,v(1≤u,v≤n) , which means there is an edge between u and v.
For the second testcase, One of the path is 1->2->3
If you doesn't know what is Hamiltonian path, click here (https://en.wikipedia.org/wiki/Hamiltonian_path).
#include<bits/stdc++.h>
using namespace std;
const int maxn=2100;
const int INF=0x3f3f3f3f;
int degree[maxn];
int vis[maxn],gra[maxn][maxn];
vector<int>G[maxn];
int n;
bool dfs(int u,int fa,int cn){
if(cn==n){
return true;
}
for(int i=0;i<G[u].size();i++){
int v=G[u][i];
if(vis[v]||v==fa){
continue;
}
vis[v]=1;
if(dfs(v,u,cn+1))
return true;
vis[v]=0; //如果没有这句,会过不了这个样例。5 2 3 2 4 4 1 1 2 5 4
}
return false;
}
void init(){//以后尽量放在前面情况,不装B
for(int i=0;i<=n+2;i++)
G[i].clear();
memset(degree,0,sizeof(degree));
memset(vis,0,sizeof(vis));
memset(gra,0,sizeof(gra));
}
int main(){
int a,b;
while(scanf("%d",&n)!=EOF){
init();
for(int i=0;i<n;i++){
scanf("%d%d",&a,&b);
if(gra[a][b]==1||a==b)
continue;
gra[a][b]=gra[b][a]=1;
G[a].push_back(b);
G[b].push_back(a);
degree[a]++,degree[b]++;
}
int deg1=0,idx=1;
for(int i=1;i<=n;i++){
if(degree[i]==1){
deg1++;
idx=i;
}
}
if(deg1>2){ //度为1的大于2个,必然不行
puts("NO");
continue;
}
vis[idx]=1;
if(dfs(idx,0,1))
puts("YES");
else puts("NO");
}
return 0;
}
HDU 5424——Rikka with Graph II——————【哈密顿路径】的更多相关文章
- hdu 5424 Rikka with Graph II(dfs+哈密顿路径)
Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so h ...
- hdu 5424 Rikka with Graph II (BestCoder Round #53 (div.2))(哈密顿通路判断)
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 ...
- HDU 5424 Rikka with Graph II
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 ...
- HDU 5831 Rikka with Parenthesis II(六花与括号II)
31 Rikka with Parenthesis II (六花与括号II) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536 ...
- HDU 5831 Rikka with Parenthesis II (栈+模拟)
Rikka with Parenthesis II 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5831 Description As we kno ...
- hdu 5831 Rikka with Parenthesis II 线段树
Rikka with Parenthesis II 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5831 Description As we kno ...
- HDU 5631 Rikka with Graph 暴力 并查集
Rikka with Graph 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5631 Description As we know, Rikka ...
- HDU 5422 Rikka with Graph
Rikka with Graph Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) ...
- HDU 6090 Rikka with Graph
Rikka with Graph 思路: 官方题解: 代码: #include<bits/stdc++.h> using namespace std; #define ll long lo ...
随机推荐
- c++ 委托构造函数
#include<iostream> ; using namespace std; class Cbox{ int a ; int b ; int c ; public: int g ; ...
- 工作中用的cobbler命令行
在使用cobbler服务器,从pxe启动虚机的时候,经常用到的cobbler命令行 1.查看注册信息 cobbler system report --name=test25 2.注册信息 cobble ...
- vue_cli下开发一个简单的模块权限系统之展现数据
这个页面是用户列表:userList就是第二张截图中的data里面的userList vue中只要改变存放数据的载体就会实现页面改变,mounted的意思是页面加载时执行这里面的函数,我们需要在页面加 ...
- linux线程池
typedef struct task_node { void *arg; /* fun arg. */ void *(*fun) (void *); /* the real work of the ...
- 牛客寒假算法基础集训营5 J 炫酷数学
链接:https://ac.nowcoder.com/acm/contest/331/J来源:牛客网 小希最近想知道一个东西,就是A+B=A|B(其中|为按位或)的二元组有多少个. 当然,直接做这个式 ...
- 洛谷P1894 [USACO4.2]完美的牛栏The Perfect Stall
题目描述 农夫约翰上个星期刚刚建好了他的新牛棚,他使用了最新的挤奶技术.不幸的是,由于工程问题,每个牛栏都不一样.第一个星期,农夫约翰随便地让奶牛们进入牛栏,但是问题很快地显露出来:每头奶牛都只愿意在 ...
- 整合feign过程中出现问题:
一:编译器报错 这个地方是由于没有被spring管理,编译器报错,可以选择加上@comoponent这个注解 也可以选择不加,启动程序是不会报错的. 二 报错为空: org.springframewo ...
- 那些年坑爹的JS题目
真是让人疑惑的基础.又是一堆奇怪的题目. 题目一. 应该是关于作用域的 function test(n) { this.x = n; return this; } var x = test(1); v ...
- Luogu P4404 [JSOI2010]缓存交换 优先队列
细节题?...调了半天.... 可以发现,每一次从缓存中删除的主存一定是下次访问最晚的,可以用优先队列来处理...还有要离散化...还有链表末尾要多建一些点...否则会死的很惨... #include ...
- hdu1430 魔板(康拓展开 bfs预处理)
魔板 Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submiss ...