Vertex Cover

frog has a graph with nn vertices v(1),v(2),…,v(n)v(1),v(2),…,v(n) and mm edges (v(a1),v(b1)),(v(a2),v(b2)),…,(v(am),v(bm))(v(a1),v(b1)),(v(a2),v(b2)),…,(v(am),v(bm)). She would like to color some vertices so that each edge has at least one colored vertex. Find the minimum number of colored vertices.

Input

The input consists of multiple tests. For each test: The first line contains 22 integers n,mn,m (2≤n≤500,1≤m≤n(n−1)22≤n≤500,1≤m≤n(n−1)2). Each of the following mm lines contains 22 integers ai,biai,bi (1≤ai,bi≤n,ai≠bi,min{ai,bi}≤301≤ai,bi≤n,ai≠bi,min{ai,bi}≤30)

Output

For each test, write 11 integer which denotes the minimum number of colored vertices.

Sample Input

3 2 1 2 1 3 6 5 1 2 1 3 1 4 2 5 2 6

Sample Output

1 2

题意：有n个点（n<=500），m条边，所有大于30的点都会与前30中的一个相连。。然后给出所有相连的点，求染多少次能让所有的点都染上

爆搜到30就行了

爆搜+减zhi（感觉也没咋减）。。。

#include<iostream> #include<algorithm> #include<queue> #include<vector> #include<cstring> #include<set> using namespace std; #define inf 0x3f3f3f3f vector<int>q[510]; int n,m; int max_ran; int ans; int vis[510]; void dfs(int k,int sum) { if(sum>ans) return; if(k>max_ran) { ans=sum; return; } if(vis[k]) dfs(k+1,sum); else { vis[k]++; dfs(k+1,sum+1); vis[k]--; int size=q[k].size(); for(int i=0;i<size;i++) { if(!vis[q[k][i]]) sum++; vis[q[k][i]]++; } dfs(k+1,sum); for(int i=0;i<size;i++) { vis[q[k][i]]--; if(!vis[q[k][i]]) sum--; } } } int main() { while(cin>>n>>m) { for(int i=0;i<=n;i++) q[i].clear(); for(int i=0;i<m;i++) { int u,v; cin>>u>>v; q[u].push_back(v); q[v].push_back(u); } ans=inf; max_ran=min(n,30); memset(vis,0,sizeof(vis)); dfs(1,0); cout<<ans<<endl; } return 0; }