HDU 4337 King Arthur's Knights(暴力dfs)
来源:互联网 发布:anaconda linux 编辑:程序博客网 时间:2024/03/29 06:35
题意:图中有N个点,每个点至少和其他(N+1)/2个点连接,问能否找到一个有N个点的环(即不存在重复的点,N个点都在内)。
分析:数据量小,直接暴力深搜,起点直接设为点1(一定能找到环 或者 环本身不存在)。
#include <iostream>#include <algorithm>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <vector>#include <set>#include <queue>#include <stack>#include <climits>//形如INT_MAX一类的#define MAX 200#define INF 0x7FFFFFFF# define eps 1e-5//#pragma comment(linker, "/STACK:36777216") ///传说中的外挂using namespace std;int map[MAX][MAX];int path[MAX];int vis[MAX];int n,m,ok;void init(){ memset(vis,0,sizeof(vis)); memset(path,0,sizeof(path));}void dfs(int u0,int v,int cur){ path[cur] = v; if(cur == n && map[u0][v]) { ok = 1; return ; } for(int i=1; i<=n; i++) { if(vis[i] == 0 && map[v][i]) { vis[i] = 1; dfs(u0,i,cur+1); vis[i] = 0 ; } if(ok == 1) return ; }}int main(){ while(cin >> n >> m) { int a,b; memset(map,0,sizeof(map)); for(int i=0; i<m; i++) { scanf("%d%d",&a,&b); map[a][b] = 1; map[b][a] = 1; } ok = 0; init(); vis[1] = 1; dfs(1,1,1); if(ok == 1) { for(int i=1; i<n; i++) { cout << path[i]<< ' '; } cout << path[n] << endl; } else cout << "no solution" << endl; } return 0;}
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