POJ 3740 DLX

题意：给你一个01矩阵，然后求是否存在选择一些行，使得每一列的1的个数都为1。

思路：貌似朴素的DFS也可以，加点剪枝就可以过。这里贴个DLX的模版。

推荐博客：http://www.cppblog.com/notonlysuccess/archive/2009/07/10/89701.html

这里讲的很详细。

 

#include <set>

#include <map>

#include <stack>

#include <cmath>

#include <queue>

#include <cstdio>

#include <string>

#include <vector>

#include <iomanip>

#include <cstring>

#include <iostream>

#include <algorithm>

#define Max 2505

#define FI first

#define SE second

#define ll long long

#define PI acos(-1.0)

#define inf 0x3fffffff

#define LL(x) ( x << 1 )

#define bug puts("here")

#define PII pair<int,int>

#define RR(x) ( x << 1 | 1 )

#define mp(a,b) make_pair(a,b)

#define mem(a,b) memset(a,b,sizeof(a))

#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )



using namespace std;



#define N 5555



///DLX

int L[N] , R[N] , D[N] , U[N] ,S[N] , C[N] ,st[N] ;//S[] 表示这一列的点数。C[] 表示这个点位于那一列。

int n , m , num , head ;



void insert(int col , int pos){//在这一列插入序号为pos的点

    int now = col ;

    while(D[now] != col) now = D[now] ;

    D[now] = pos ;

    D[pos] = col ;

    U[pos] = now ;

    U[col] = pos ;

}



void init(){

    head = 0 ;

    R[head] = 1 ;L[head] = m ;

    for (int i = 1 ; i <= m ; i ++ ){//每一行的头指针

        if(i == 1)L[i] = head ;

        else L[i] = i - 1 ;

        if(i == m)R[i] = head ;

        else R[i] = i + 1 ;

        U[i] = i ;

        D[i] = i ;

        S[i] = 0 ;

        C[i] = i ;

    }

    num = m ;//已经插入m个节点

    int k ;

    for (int i = 1 ; i <= n ; i ++ ){

        mem(st ,0) ;

        for (int j = 1 ; j <= m ; j ++ ){

            scanf("%d",&k) ;

            if(!k)continue ;

            num ++ ;

            insert(j , num) ;

            if(st[0] == 0){//每行的第一个

                L[num] = num ; R[num] = num ;

            }else{

                L[num] = st[st[0]] ;

                R[num] = st[1] ;

                R[st[st[0]]] = num ;

                L[st[1]] = num ;

            }

            st[++st[0]] = num ;

            C[num] = j ;

            S[j] ++ ;

        }

    }

}



void remove(const int &c){//删除

    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[C[j]] ;

        }

    }

}



void resume(const int &c){//恢复

    for (int i = U[c] ; i != c ; i = U[i]){

        for (int j = L[i] ; j != i ; j = L[j]){

            ++ S[C[j]] ;

            U[D[j]] = j ;

            D[U[j]] = j ;

        }

    }

    L[R[c]] = c ;

    R[L[c]] = c ;

}



int dfs(const int &k){

    if(R[head] == head)return 1 ;

    int MX = inf ,c ;

    for (int t = R[head] ; t != head ; t = R[t]){//找出点最少的一列

        if(S[t] < MX){

            MX = S[t] ;

            c = t ;

        }

    }

    remove(c) ;

    for (int i = D[c] ; i != c ; i = D[i]){

        for (int j = R[i] ; j != i ; j = R[j]){

            remove(C[j]) ;

        }

        if(dfs(k + 1))return 1 ;

        for (int j = L[i] ; j != i ; j = L[j]){

            resume(C[j]) ;

        }

    }

    resume(c) ;

    return 0 ;

}

int main() {

    while(cin >> n >> m){

        init() ;

        if(dfs(0))puts("Yes, I found it") ;

        else puts("It is impossible") ;

    }

    return 0 ;

}