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Rikka with Graph II

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?

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.

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

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).

哈密顿通路：经过每个点有且仅有一次的一条通路。

由于图，只有n个边，那么最小度的那个边，最大只有2，先判定，是否连通，如果连通，每次，都找那个度最小的作为入点，因为，最小度的点，一定会成为某条路的端点，这样就可以用o(n )的复杂度，找到一条哈密顿回路。

给个测试数据，

5 

1 5 1 2 2 3 2 4 1 3

#define N 1005
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,a,b,d[N],minx,mi;
vector<int> p[N];
bool vis[N];
void DFS(int x){
    vis[x] = true;
    FI(p[x].size()){
        int goal = p[x][i];
        if(!vis[goal]){
            DFS(goal);
        }
    }
}
int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
     while(S(n)!=EOF)
    {
        FI(n+1) p[i].clear();
        fill(d,0);
        FI(n){
            S2(a,b);
            if(a != b){
               p[a].push_back(b);
               p[b].push_back(a);
               d[a]++;d[b]++;
            }
        }
        bool isConnect = true;
        fill(vis,false);
        DFS(1);
        For(i,1,n+1){
            if(!vis[i]){
                isConnect = false;
                break;
            }
        }
        if(!isConnect){
            printf("NO\n");
            continue;
        }
        minx = N;mi = 1;
        For(i,1,n+1){
            if(d[i] < minx){
                mi = i;
                minx = d[i];
            }
        }
        fill(vis,false);
        vis[mi] = true;
        while(true){
            bool isChange = false;
            minx = N;
            int mit = -1;
            FI(p[mi].size()){
                int goal = p[mi][i];
                if(!vis[goal]){
                    if(d[goal] < minx){
                        mit = goal;
                        minx = d[goal];
                    }
                    isChange = true;
                }
            }
            if(!isChange) break;
            mi = mit;
            vis[mi] = true;
        }
        isConnect = true;
        For(i,1,n+1){
            if(!vis[i]){
                isConnect = false;
                break;
            }
        }
        if(!isConnect)
            printf("NO\n");
        else
            printf("YES\n");
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}

hdu 5424 Rikka with Graph II 哈密顿通路

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原文地址：http://blog.csdn.net/mengzhengnan/article/details/48093985