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///DLXint 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 ;}