POJ 3740 Easy Finding DLX

题意：给你一个0,1矩阵 ，求精确覆盖

解题思路：

DLX

解题代码：

  1 // File Name: poj3740.cpp
  2 // Author: darkdream
  3 // Created Time: 2014年10月04日 星期六 20时06分31秒
  4
  5 #include<vector>
  6 #include<list>
  7 #include<map>
  8 #include<set>
  9 #include<deque>
 10 #include<stack>
 11 #include<bitset>
 12 #include<algorithm>
 13 #include<functional>
 14 #include<numeric>
 15 #include<utility>
 16 #include<sstream>
 17 #include<iostream>
 18 #include<iomanip>
 19 #include<cstdio>
 20 #include<cmath>
 21 #include<cstdlib>
 22 #include<cstring>
 23 #include<ctime>
 24 #define LL long long
 25
 26 using namespace std;
 27 const int maxnode = 50000;
 28 const int MaxN = 18;
 29 const int MaxM = 400;
 30 struct DLX
 31 {
 32     int n , m , size;
 33     int U[maxnode],D[maxnode],L[maxnode],R[maxnode],Row[maxnode],Col[maxnode];
 34     int H[MaxN],S[MaxM];
 35     void init(int _n,int _m)
 36     {
 37        n = _n;
 38        m = _m;
 39        for(int i = 0 ;i <= m;i ++)
 40        {
 41            S[i] = 0 ;
 42            U[i] = i ;
 43            D[i] = i ;
 44            L[i] = i-1;
 45            R[i] = i+1;
 46        }
 47        size = m ;
 48        L[0] = m;
 49        R[m] = 0 ;
 50        memset(H,-1,sizeof(H));
 51     }
 52     void Link(int r, int c)
 53     {
 54         ++S[Col[++size] = c];
 55         Row[size] = r;
 56         D[size] = D[c];
 57         U[D[c]] = size;
 58         U[size] = c;
 59         D[c] = size ;
 60         if(H[r] < 0)  H[r] = L[size] = R[size] = size;
 61         else{
 62           R[size] = R[H[r]];
 63           L[R[H[r]]] = size;
 64           L[size] = H[r];
 65           R[H[r]] = size;
 66         }
 67     }
 68     void remove(int c)
 69     {
 70       L[R[c]] = L[c];
 71       R[L[c]] = R[c];
 72       for(int i = D[c]; i != c ;i = D[i])
 73           for(int j = R[i]; j != i ;j = R[j])
 74           {
 75                U[D[j]] = U[j];
 76                D[U[j]] = D[j];
 77                --S[Col[j]];
 78           }
 79     }
 80     void resume(int c)
 81     {
 82        L[R[c]] = c;
 83        R[L[c]] = c;
 84        for(int i = D[c];i != c;i = D[i])
 85            for(int j = R[i]; j != i ;j = R[j])
 86            {
 87               U[D[j]] = j ;
 88               D[U[j]] = j;
 89               ++ S[Col[j]];
 90            }
 91     }
 92     bool Dance()
 93     {
 94        if(R[0] == 0 )
 95        {
 96          return true;
 97        }
 98        int c = R[0];
 99        for(int i = R[c];i != 0 ;i = R[i])
100            if(S[i] < S[c])
101                c = i ;
102        remove(c);
103        for(int i = D[c];i != c;i = D[i])
104        {
105            for(int j = R[i];j != i ;j = R[j]) remove(Col[j]);
106            if(Dance()) return true;
107            for(int j = R[i];j != i; j = R[j]) resume(Col[j]);
108        }
109        resume(c);
110        return false;
111     }
112 };
113
114 DLX g;
115 int main(){
116     int n , m ;
117     while(scanf("%d %d",&n,&m) != EOF)
118     {
119        g.init(n,m);
120        int t ;
121        for(int i = 1;i <= n;i ++)
122            for(int j = 1;j <= m;j ++)
123            {
124               scanf("%d",&t);
125               if(t)
126                 g.Link(i,j);
127            }
128        if(g.Dance())
129            printf("Yes, I found it\n");
130        else printf("It is impossible\n");
131     }
132 return 0;
133 }