hdu 4337 （哈密顿回路模板）

King Arthur's Knights

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1557    Accepted Submission(s): 661
Special Judge

Problem Description

I am the bone of my sword. Steel is my body, and the fire is my blood.
- from Fate / Stay Night
You must have known the legend of King Arthur and his knights of the round table. The round table has no head, implying that everyone has equal status. Some knights are close friends with each other, so they prefer to sit next to each other.

Given the relationship of these knights, the King Arthur request you to find an arrangement such that, for every knight, his two adjacent knights are both his close friends. And you should note that because the knights are very united, everyone has at least half of the group as his close friends. More specifically speaking, if there are N knights in total, every knight has at least (N + 1) / 2 other knights as his close friends.

 

Input

The first line of each test case contains two integers N (3 <= N <= 150) and M, indicating that there are N knights and M relationships in total. Then M lines followed, each of which contains two integers ai and bi (1 <= ai, bi <= n, ai != bi), indicating that knight ai and knight bi are close friends.

 

Output

For each test case, output one line containing N integers X1, X2, ..., XN separated by spaces, which indicating an round table arrangement. Please note that XN and X1 are also considered adjacent. The answer may be not unique, and any correct answer will be OK. If there is no solution exists, just output "no solution".

 

Sample Input

  

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

 

Sample Output

  

   1 2 3
1 4 2 3
  

 

Source

2012 Multi-University Training Contest 4

 

Recommend

zhoujiaqi2010

 

这个题很容易看出是找一个哈密顿回路，关键是哈密顿回路怎么求？其实我不会，直接找模板，结果就搜到这题的解题报告了，囧，可见哈密顿回路的题还是挺少的。这题比较特殊的是题中给了条件every knight has at least (N + 1) / 2 other knights as his close friends.这句话其实是保证了哈密顿回路的存在性，所以不会有no solution。http://www.cnblogs.com/markliu/archive/2012/08/03/2621906.html代码稍加修改，模板留存。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <map>
#include <vector>
#include<queue>
#include<set>
#include<cmath>
using namespace std;
typedef long long LL;
typedef pair<double,int> P;
const int maxn = 200 + 5;
const int INF = 1000000000;

int n,m;
bool G[maxn][maxn];
int x[maxn];//记录回路中的第i个点
//bool flag[maxn];

void hamilton(){
    int k;
    bool s[maxn];
    for(int i = 0; i < n; i++){
        x[i] = -1;
        s[i] = false;
    }
    k = 1;
    s[0] = true;
    x[0] = 0;
    while(k >= 0){
        x[k]++;
        while(x[k] < n){
            if(!s[x[k]] && G[x[k - 1]][x[k]])
                break;
            else
                x[k]++;
        }
        if((x[k] < n) && (k != n - 1)){
            s[x[k++]] = true;
        }
        else if((x[k] < n) && k == n - 1 && G[x[k]][x[0]])
            break;
        else{
            x[k] = -1;
            k--;
            s[x[k]] = false;
        }
    }
}

int main(){
    while(scanf("%d%d",&n,&m) != EOF){
        memset(G,0,sizeof(G));
        //memset(flag,0,sizeof(flag));
        memset(x,0,sizeof(x));
        for(int i = 0;i < m;i++){
            int a,b;
            scanf("%d%d",&a,&b);a--;b--;//hamilton函数中下标从0开始
            G[a][b]=G[b][a]=true;
        }

        hamilton();
        /*bool f=0;
        for(int i=0;i<n;i++){
            flag[x[i]]=1;
        }
        for(int i=0;i<n;i++){
            if(!flag[i]) f=1;
        }
        if(f) cout <<"no solution" <<endl;*/
        //else{
        for(int i = 0;i < n;i++){
            if(i == n-1) printf("%d\n",x[i]+1);
            else printf("%d ",x[i]+1);
        }
        //}
    }
    return 0;
}