POJ3740--Easy Finding(Dancing Links)

Description

Given a  M× N matrix  A.  A _ij ∈ {0, 1} (0 ≤ i < M, 0 ≤ j < N), could you find some rows that let every cloumn contains and only contains one 1.

Input

There are multiple cases ended by  EOF. Test case up to 500.The first line of input is  M,  N ( M ≤ 16,  N ≤ 300). The next  M lines every line contains  N integers separated by space.

Output

For each test case, if you could find it output "Yes, I found it", otherwise output "It is impossible" per line.

Sample Input

3 3
0 1 0
0 0 1
1 0 0
4 4
0 0 0 1
1 0 0 0
1 1 0 1
0 1 0 0

Sample Output

Yes, I found it
It is impossible

思路：精确覆盖基础题，用Dancing Links。不懂谷歌knuth的X算法。。模板是copy来的=_=

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int maxnode = 100010;
const int MaxM = 1010;
const int MaxN = 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];//本题的话这两个记录最小值和记录选择的数组不需要、
    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)
    {
        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]];
        L[R[c]] = R[L[c]] = c;
    }
    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) == 2)
    {
        g.init(n,m);
        for(int i = 1;i <= n;i++)
        {
			for(int j = 1;j <= m;j++)
			{
				int a;
				scanf("%d",&a);
				if(a)	g.Link(i,j);
			}
        }
        if(!g.Dance(0))printf("It is impossible\n");
        else printf("Yes, I found it\n");
    }
    return 0;
}