【DLX算法】hdu3498 whosyourdaddy

题意：给你一个01矩阵，让你选择尽可能少的行数，使得这些行的并集能够覆盖到所有列。

DLX算法求解重复覆盖问题模板，使用估价函数进行剪枝。

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=55+5;
const int maxr=55+5;
const int maxnode=55*55+5;
//ÐбàºÅ´Ó1¿ªÊ¼£¬ÁбàºÅΪ1~n£¬½áµã0ÊDZíÍ·½áµã£»½áµã1~nÊǸ÷Áж¥²¿µÄÐéÄâ½áµã 
struct DLX{
	int n,sz;//ÁÐÊý£¬½áµã×ÜÊý 
	int S[maxn];//¸÷ÁнáµãÊý 
	int row[maxnode],col[maxnode];//¸÷½áµãËùÔÚµÄÐÐÁбàºÅ 
	int L[maxnode],R[maxnode],U[maxnode],D[maxnode];
	int ansd,ans[maxr];//½â 
	void init(int n){//nÊÇÁÐÊý 
		this->n=n;
		for(int i=0;i<=n;++i){
			U[i]=i;
			D[i]=i;
			L[i]=i-1;
			R[i]=i+1;
		}
		R[n]=0; L[0]=n;
		sz=n+1; ansd=1000000007;
		memset(S,0,sizeof(S));
	}
	void addRow(int r,vector<int> columns){
		int first=sz;
		for(int i=0;i<columns.size();++i){
			int c=columns[i];
			L[sz]=sz-1;
			R[sz]=sz+1;
			D[sz]=c;
			U[sz]=U[c];
			D[U[c]]=sz;
			U[c]=sz;
			row[sz]=r;
			col[sz]=c;
			++S[c];
			++sz;
		}
		R[sz-1]=first;
		L[first]=sz-1;
	}
	//˳×ÅÁ´±íA£¬±éÀú³ýsÍâµÄÆäËûÔªËØ 
	#define FOR(i,A,s) for(int i=A[s];i!=s;i=A[i])
	void remove(int c){
		L[R[c]]=L[c];
		R[L[c]]=R[c];
		FOR(i,D,c){
			FOR(j,R,i){
				U[D[j]]=U[j];
				D[U[j]]=D[j];
				--S[col[j]];
			}
		}
	}
	void restore(int c){
		FOR(i,U,c){
			FOR(j,L,i){
				++S[col[j]];
				U[D[j]]=j;
				D[U[j]]=j;
			}
		}
		L[R[c]]=c;
		R[L[c]]=c;
	}
	bool dfs(int d){
//		printf("%d",d);
		if(R[0]==0){//ÕÒµ½½â 
			ansd=d;//¼Ç¼½âµÄ³¤¶È 
			return 1;
		}
		//ÕÒ½áµãÊý×îСµÄÁÐc 
		int c=R[0];//µÚÒ»¸öδɾ³ýµÄÁÐ 
		FOR(i,R,0){
			if(S[i]<S[c]){
				c=i;
			}
		}
		remove(c);//ɾ³ýµÚcÁÐ 
		FOR(i,D,c){//ÓýáµãiËùÔÚÐи²¸ÇµÚcÁÐ 
			ans[d]=row[i];
			FOR(j,R,i){
				remove(col[j]);//ɾ³ý½áµãiËùÔÚÐÐÄܸ²¸ÇµÄËùÓÐÆäËûÁÐ 
			}
			if(dfs(d+1)){
				return 1;
			}
			FOR(j,L,i){
				restore(col[j]);//»Ö¸´½áµãiËùÔÚÐÐÄܸ²¸ÇµÄÆäËûËùÓÐÁÐ 
			}
		}
		restore(c);//»Ö¸´µÚcÁÐ 
		return 0;
	}
	bool solve(vector<int>& v){
		v.clear();
		if(!dfs(0)){
			return 0;
		}
		for(int i=0;i<ansd;++i){
			v.push_back(ans[i]);
		}
		return 1;
	}
	
	void remov2(int c){
		FOR(i,D,c){
			L[R[i]]=L[i];
			R[L[i]]=R[i];
		}
    }
    void restor2(int c){
		FOR(i,U,c){
			L[R[i]]=R[L[i]]=i;
		}
    }
    int f(){//¹À¼Ûº¯Êý 
    	bool vis[maxn];
    	int res=0;
    	FOR(i,R,0){
    		vis[i]=1;
    	}
    	FOR(i,R,0){
    		if(vis[i]){
    			++res;
    			vis[i]=0;
    			FOR(j,D,i){
    				FOR(k,R,j){
    					vis[col[k]]=0;
    				}
    			}
    		}
    	}
    	return res;
    }
    void df2(int d){
    	if(d+f()>=ansd){
    		return;
    	}
		if(R[0]==0){//ÕÒµ½½â 
			ansd=min(ansd,d);//¼Ç¼½âµÄ³¤¶È 
			return;
		}
		//ÕÒ½áµãÊý×îСµÄÁÐc 
		int c=R[0];//µÚÒ»¸öδɾ³ýµÄÁÐ 
		FOR(i,R,0){
			if(S[i]<S[c]){
				c=i;
			}
		}
		FOR(i,D,c){
			remov2(i);
			FOR(j,R,i){
				remov2(j);
			}
			df2(d+1);
			FOR(j,L,i){
				restor2(j);
			}
			restor2(i);
		}
	}
    
}dlx;
int n,m;
bool a[60][60];
int main(){
	//freopen("hdu3498.in","r",stdin);
	int x,y;
	while(scanf("%d%d",&n,&m)!=EOF){
		dlx.init(n);
		memset(a,0,sizeof(a));
		for(int i=1;i<=n;++i){
			a[i][i]=1;
		}
		for(int i=1;i<=m;++i){
			scanf("%d%d",&x,&y);
			a[x][y]=a[y][x]=1;
		}
		for(int i=1;i<=n;++i){
			vector<int> columns;
			for(int j=1;j<=n;++j){
				if(a[i][j]){
					columns.push_back(j);
				}
			}
			if(!columns.empty()){
				dlx.addRow(i,columns);
			}
		}
		dlx.df2(0);
		printf("%d
",dlx.ansd);
	}
	return 0;
}