BestCoder Round #53 (div.2) HDOJ5424 Rikka with Graph II(dfs & 哈密顿通路判断)

Rikka with Graph II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 717    Accepted Submission(s): 174

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 if there exist a Hamiltonian path.

It is too difficult for Rikka. Can you help her?

 

Input

There are no more than 100 testcases. 

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.

 

Output

For each testcase, if there exist a Hamiltonian path print "YES" , otherwise print "NO".

 

Sample Input

4
1 1
1 2
2 3
2 4
3
1 2
2 3
3 1

 

Sample Output

NO
YES

Hint
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).

 

问你一个图是否存在哈密顿通路。

哈密顿通路：经过图中每个点一次的一条通路。

直接套板子必然会TLE，加用dfs优化。先判断图是否连通，连通则每次找度最小的点开始。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int MAXN = 1005;
int n, cnt, deg[MAXN];
bool flag, vis[MAXN], map[MAXN][MAXN];
void dfs(int x)
{
	if(cnt == n) {
		flag = true;
		return ;
	}
	for(int i = 1; i <= n && !flag; ++i)
		if(!vis[i] && map[x][i]) {
			vis[i] = true;
			++cnt;
			dfs(i);
			--cnt;
			vis[i] = false;
		}
}
int main(int argc, char const *argv[])
{
	while(scanf("%d", &n) != EOF) {
		memset(deg, 0, sizeof(deg));
		memset(vis, false, sizeof(vis));
		memset(map, false, sizeof(map));
		flag = false;
		cnt = 1;
		for(int i = 0; i < n; ++i) {
			int x, y;
			scanf("%d%d", &x, &y);
			if(x != y && !map[x][y]) {
				map[x][y] = map[y][x] = true;
				++deg[x], ++deg[y];
			}
		}
		int tmp = 1, sum = 0;
		for(int i = 1; i <= n; ++i)
			if(deg[i] == 1) {
				tmp = i;
				++sum;
			}
		if(sum > 2) {
			printf("NO\n");
			continue;
		}
		vis[tmp] = true;
		dfs(tmp);
		if(flag) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}