HUST 1017(DLX)

题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=65998#problem/A

题意:求01矩阵的精确覆盖。

DLX学习资料:http://www.cnblogs.com/grenet/p/3145800.html

                    http://blog.csdn.net/mu399/article/details/7627862

#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstdlib>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define LL long long
#define mod 10007
#define inf 0x3f3f3f3f
#define N 100010
#define FILL(a,b) (memset(a,b,sizeof(a)))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int maxnode=100010;
const int MaxN=1010;
const int MaxM=1010;
struct DLX
{
    int n,m,size;
    int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];
    int  H[MaxN],S[MaxM];
    int ansd,ans[MaxN];
    void init(int _n,int _m)
    {
        n=_n;m=_m;
        for(int i=0;i<=m;i++)
        {
            S[i]=0;
            U[i]=D[i]=i;
            L[i]=i-1;
            R[i]=i+1;
        }
        R[m]=0;L[0]=m;
        size=m;
        for(int i=1;i<=n;i++)H[i]=-1;
    }
    void Link(int r,int c)
    {
        ++S[Col[++size]=c];
        Row[size]=r;
        D[size]=D[c];
        U[D[c]]=size;
        U[size]=c;
        D[c]=size;
        if(H[r]<0)H[r]=L[size]=R[size]=size;
        else
        {
            R[size]=R[H[r]];
            L[R[H[r]]]=size;
            L[size]=H[r];
            R[H[r]]=size;
        }
    }
    void Remove(int c)
    {
        L[R[c]]=L[c];R[L[c]]=R[c];
        for(int i=D[c];i!=c;i=D[i])
            for(int j=R[i];j!=i;j=R[j])
        {
            U[D[j]]=U[j];
            D[U[j]]=D[j];
            --S[Col[j]];
        }
    }
    void Resume(int c)
    {
        L[R[c]]=R[L[c]]=c;
        for(int i=U[c];i!=c;i=U[i])
            for(int j=L[i];j!=i;j=L[j])
            ++S[Col[U[D[j]]=D[U[j]]=j]];
    }
    bool Dance(int d)
    {
        if(R[0]==0)
        {
            ansd=d;
            return true;
        }
        int c=R[0];
        for(int i=R[0];i!=0;i=R[i])
            if(S[i]<S[c])c=i;
        Remove(c);
        for(int i=D[c];i!=c;i=D[i])
        {
            ans[d]=Row[i];
            for(int j=R[i];j!=i;j=R[j])Remove(Col[j]);
            if(Dance(d+1))return true;
            for(int j=L[i];j!=i;j=L[j])Resume(Col[j]);
        }
        Resume(c);
        return false;
    }
};
DLX g;
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)>0)
    {
        g.init(n,m);
        for(int i=1;i<=n;i++)
        {
            int num,j;
            scanf("%d",&num);
            while(num--)
            {
                scanf("%d",&j);
                g.Link(i,j);
            }
        }
        if(!g.Dance(0))puts("NO");
        else
        {
            printf("%d",g.ansd);
            for(int i=0;i<g.ansd;i++)printf(" %d",g.ans[i]);
            puts("");

        }
    }
}
View Code
原文地址:https://www.cnblogs.com/lienus/p/4245338.html