AcWing 1067. 精确覆盖问题 DLX

地址 https://www.acwing.com/problem/content/description/1069/

给定一个 N×M 的数字矩阵 A，矩阵中的元素 Ai,j∈{0,1}。

请问，你能否在矩阵中找到一个行的集合，使得这些行中，每一列都有且仅有一个数字 1。

输入格式
第一行包含两个整数 N 和 M。

接下来 N 行，每行包含 M 个整数（0 或 1），表示完整的数字矩阵。

输出格式
如果能找到满足条件的行的集合，则在一行中依次输出这些行的编号（行编号 1∼N)。

如果方案不唯一，则以任意顺序输出任意方案即可。

否则，输出 No Solution!。

数据范围
1≤N,M≤500,
数据保证矩阵中 1 的数量不超过 5000。

输入样例1：
3 3
0 1 0
0 0 1
1 0 0
输出样例1：
1 2 3
输入样例2：
4 4
0 0 0 1
1 0 0 0
1 1 0 1
0 1 0 0
输出样例2：
No Solution!

自写指针版本 DLX
这里是自己整理重新画图的V2版本 DLX学习心得

// DLX_2.cpp : 此文件包含 "main" 函数。程序执行将在此处开始并结束。
//

#include <iostream>
#include <vector>
#include <assert.h>

using namespace std;

const int N = 510;


/*
3 3
0 1 0
0 0 1
1 0 0

1 2 3

4 4
0 0 0 1
1 0 0 0
1 1 0 1
0 1 0 0

No Solution!
*/


struct DLX_NODE {
    struct DLX_NODE* up;
    struct DLX_NODE* down;
    struct DLX_NODE* left;
    struct DLX_NODE* right;
    int row; int col;
    int count;
    int ele;

    DLX_NODE() {
        up = NULL; down = NULL; left = NULL; right = NULL;
        col = -1; row = -1; ele = -1; count = 0;
    }
};

struct DLX_NODE table[N][N];

int n, m;

void init()
{
    for (int i = 0; i < n; i++) {
        table[i][0].down = &table[i + 1][0];
        table[i + 1][0].up = &table[i][0];
        table[i][0].row = i; table[i][0].col = 0;
        table[i + 1][0].row = i + 1; table[i + 1][0].col = 0;
    }

    for (int i = 0; i < m; i++) {
        table[0][i].right = &table[0][i + 1];
        table[0][i + 1].left = &table[0][i];
        table[0][i].row = 0; table[0][i].col = i;
        table[0][i+1].row = 0; table[0][i+1].col = i+1;
    }


}

void printDLX()
{
    struct DLX_NODE* p = &table[0][0];
    cout << "=================================================>
";
    p = p->right;
    while (p != NULL) {
        struct DLX_NODE* pp = p;
        if (pp->count != 0) {
            cout << "colBase.count = " << pp->count << endl;
        }
        while (pp != NULL) {
            if (pp->ele == 1)
                cout << "pp->col = " << pp->col << ". pp->row = " << pp->row << ".pp->ele = " << pp->ele << endl;
            pp = pp->down;
        }
        p = p->right;
    }
    cout << endl << endl;
    p = &table[0][0];
    p = p->down;
    while (p != NULL) {
        struct DLX_NODE* pp = p;
        while (pp != NULL) {
            if (pp->ele == 1)
                cout << "pp->col = " << pp->col << ". pp>row = " << pp->row << ".pp->ele = " << pp->ele << endl;
            pp = pp->right;
        }
        p = p->down;
    }
    cout << endl ;
}



void FindColInsert(int row,int col)
{
    struct DLX_NODE* p = &table[0][col];
    while (p != NULL) {
        struct DLX_NODE* next = p->down;
        //找到链表结尾或者 next比当前元素行数更大 P的下面即是插入的位置
        if (next == NULL)  break;
        else if (next->row > row) break;
        else if (next->row == row) {
            //行列都相同则说明重复插入了
            assert(0);
        }
        p = next;
    }
    assert(p != NULL);
    //插入
    struct DLX_NODE* node = &table[row][col];
    node->down = p->down;
    if (p->down != NULL) p->down->up = node;

    node->up = p;
    p->down = node;

    table[0][col].count++;
}


void FindRowInsert(int row, int col)
{
    struct DLX_NODE* p = &table[row][0];
    while (p != NULL) {
        struct DLX_NODE* next = p->right;
        //找到链表结尾或者 next比当前元素列数更大 P的右面即是插入的位置
        if (next == NULL)  break;
        else if (next->col > col) break;
        else if (next->col == col) {
            //行列都相同则说明重复插入了
            assert(0);
        }
        p = next;
    }
    assert(p != NULL);
    //插入
    struct DLX_NODE* node = &table[row][col];
    node->right = p->right;
    if (p->right != NULL) p->right->left = node;

    node->left = p;
    p->right = node;
}

void addNode(int num, int x, int y)
{
    table[x][y].row = x;
    table[x][y].col = y;
    table[x][y].ele = num;

    //插入列中
    FindColInsert(x,y);

    //插入行中
    FindRowInsert(x,y);
}

void resume(struct DLX_NODE* pToResume)
{
    struct DLX_NODE* pdown = pToResume->down;
    while (pdown != NULL) {
        int row = pdown->row;
        assert(row != -1);
        struct DLX_NODE* nextResume = &table[row][0];
        while (nextResume != NULL) {
            if (nextResume->col > 0)
                table[0][nextResume->col].count++;
            nextResume->up->down = nextResume;
            if (nextResume->down != NULL) nextResume->down->up = nextResume;
            nextResume = nextResume->right;
        }
        pdown = pdown->down;
    }


    if (pToResume->right != NULL) pToResume->right->left = pToResume;
    pToResume->left->right = pToResume;


}

void remove(struct DLX_NODE* pToRemove)
{
    if (pToRemove->right != NULL) { pToRemove->right->left = pToRemove->left; }
    pToRemove->left->right = pToRemove->right;

    struct DLX_NODE* pdown = pToRemove->down;
    while (pdown != NULL) {
        int row = pdown->row;
        assert(row != -1);
        struct DLX_NODE* nextRemove = &table[row][0];
        while (nextRemove != NULL) {
            if(nextRemove->col != 0)
                table[0][nextRemove->col].count--;
            assert(table[0][nextRemove->col].count >= 0);
            if (nextRemove->col == pToRemove->col) {
                nextRemove = nextRemove->right;
                continue;
            }
            if (nextRemove->down != NULL) nextRemove->down->up = nextRemove->up;
            nextRemove->up->down = nextRemove->down;
            nextRemove = nextRemove->right;
        }
        pdown = pdown->down;
    }
}

vector<int> ans;
bool dfs()
{
    int noEle = 1;  int minCount = 99999; struct DLX_NODE* selectP = NULL;
    struct DLX_NODE* p = table[0][0].right;
    if (p == NULL) 
        return true;
    while (p != NULL) {
        if (p->count != -1 && p->count < minCount) {
            minCount = p->count;
            selectP = p;
        }
        p = p->right;
    }


    remove(selectP);
    {
        p = selectP->down;
        while (p != NULL) {
            ans.push_back(p->row);
            struct DLX_NODE* tmp = table[p->row][0].right;
            while (tmp != NULL) {
                if(tmp->col != selectP->col)
                    remove(&table[0][tmp->col]);
                tmp = tmp->right;
            }
            if (dfs()) return true;

            tmp = table[p->row][0].right;
            while (tmp != NULL) {
                if (tmp->col != selectP->col)
                    resume(&table[0][tmp->col]);
                tmp = tmp->right;
            }
            ans.pop_back();
            p = p->down;
        }
    }

    resume(selectP);

    return false;
}



int main()
{
    cin >> n >> m;
    init();
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            int x; cin >> x;
            if (x) {
                addNode(x, i, j);
            }
        }
    }
    //printDLX();
    if (dfs())
    {
        for(auto& e:ans)  
            printf("%d ", e);
        puts("");
    }
    else puts("No Solution!");



    return 0;
}


作者：itdef
链接：https://www.acwing.com/solution/content/26693/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

View Code

实际使用建议使用Y总模板
使用了链式前向星的十字链表DLX

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 const int N = 5510;
 8 
 9 int n, m;
10 int l[N], r[N], u[N], d[N], s[N], row[N], col[N], idx;
11 int ans[N], top;
12 
13 void init()
14 {
15     for (int i = 0; i <= m; i ++ )
16     {
17         l[i] = i - 1, r[i] = i + 1;
18         u[i] = d[i] = i;
19     }
20     l[0] = m, r[m] = 0;
21     idx = m + 1;
22 }
23 
24 void add(int& hh, int& tt, int x, int y)
25 {
26     row[idx] = x, col[idx] = y, s[y] ++ ;
27     u[idx] = y, d[idx] = d[y], u[d[y]] = idx, d[y] = idx;
28     r[hh] = l[tt] = idx, r[idx] = tt, l[idx] = hh;
29     tt = idx ++ ;
30 }
31 
32 void remove(int p)
33 {
34     r[l[p]] = r[p], l[r[p]] = l[p];
35     for (int i = d[p]; i != p; i = d[i])
36         for (int j = r[i]; j != i; j = r[j])
37         {
38             s[col[j]] -- ;
39             u[d[j]] = u[j], d[u[j]] = d[j];
40         }
41 }
42 
43 void resume(int p)
44 {
45     for (int i = u[p]; i != p; i = u[i])
46         for (int j = l[i]; j != i; j = l[j])
47         {
48             u[d[j]] = j, d[u[j]] = j;
49             s[col[j]] ++ ;
50         }
51     r[l[p]] = p, l[r[p]] = p;
52 }
53 
54 bool dfs()
55 {
56     if (!r[0]) return true;
57     int p = r[0];
58     for (int i = r[0]; i; i = r[i])
59         if (s[i] < s[p])
60             p = i;
61     remove(p);
62     for (int i = d[p]; i != p; i = d[i])
63     {
64         ans[ ++ top] = row[i];
65         for (int j = r[i]; j != i; j = r[j]) remove(col[j]);
66         if (dfs()) return true;
67         for (int j = l[i]; j != i; j = l[j]) resume(col[j]);
68         top -- ;
69     }
70     resume(p);
71     return false;
72 }
73 
74 int main()
75 {
76     scanf("%d%d", &n, &m);
77     init();
78     for (int i = 1; i <= n; i ++ )
79     {
80         int hh = idx, tt = idx;
81         for (int j = 1; j <= m; j ++ )
82         {
83             int x;
84             scanf("%d", &x);
85             if (x) add(hh, tt, i, j);
86         }
87     }
88 
89     if (dfs())
90     {
91         for (int i = 1; i <= top; i ++ ) printf("%d ", ans[i]);
92         puts("");
93     }
94     else puts("No Solution!");
95 
96     return 0;
97 }

View Code

//====================================================

//数独的DLX解答代码

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 20000;

int m = 9 * 9 * 4;
int u[N], d[N], l[N], r[N], s[N], col[N], row[N], idx;
int ans[N], top;
struct Op
{
    int x, y;
    char z;
}op[N];
char g[20][9];
char input[90];

void init()
{
    for (int i = 0; i <= m; i++)
    {
        l[i] = i - 1, r[i] = i + 1;
        s[i] = 0;
        d[i] = u[i] = i;
    }
    l[0] = m, r[m] = 0;
    idx = m + 1;
}

void add(int& hh, int& tt, int x, int y)
{
    row[idx] = x, col[idx] = y, s[y] ++;
    u[idx] = y, d[idx] = d[y], u[d[y]] = idx, d[y] = idx;
    r[hh] = l[tt] = idx, r[idx] = tt, l[idx] = hh;
    tt = idx++;
}

void remove(int p)
{
    r[l[p]] = r[p], l[r[p]] = l[p];
    for (int i = d[p]; i != p; i = d[i])
        for (int j = r[i]; j != i; j = r[j])
        {
            s[col[j]] --;
            u[d[j]] = u[j], d[u[j]] = d[j];
        }
}

void resume(int p)
{
    for (int i = u[p]; i != p; i = u[i])
        for (int j = l[i]; j != i; j = l[j])
        {
            u[d[j]] = j, d[u[j]] = j;
            s[col[j]] ++;
        }
    r[l[p]] = p, l[r[p]] = p;
}

bool dfs()
{
    if (!r[0]) return true;
    int p = r[0];
    for (int i = r[0]; i; i = r[i])
        if (s[i] < s[p])
            p = i;
    remove(p);
    for (int i = d[p]; i != p; i = d[i])
    {
        ans[++top] = row[i];
        for (int j = r[i]; j != i; j = r[j]) remove(col[j]);
        if (dfs()) return true;
        for (int j = l[i]; j != i; j = l[j]) resume(col[j]);
        top--;
    }
    resume(p);
    return false;
}

int main()
{
    while (~scanf("%s", input))
    {
        if (input[0] == 'e') break;

        memset(g, 0, sizeof(g[0][0]) * 20 * 9);
        int ii = 0;
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                g[i][j] = input[ii++];
            }
        }

        init();

        for (int i = 0, n = 1; i < 9; i++)
            for (int j = 0; j < 9; j++)
            {
                int a = 0, b = 8;
                if (g[i][j] != '.') a = b = g[i][j] - '1';
                for (int k = a; k <= b; k++, n++)
                {
                    int hh = idx, tt = idx;
                    op[n] = { i, j, (char)(k + '1') };
                    add(hh, tt, n, i * 9 + j + 1);
                    add(hh, tt, n, 81 + i * 9 + k + 1);
                    add(hh, tt, n, 81 * 2 + j * 9 + k + 1);
                    add(hh, tt, n, 81 * 3 + (i / 3 * 3 + j / 3) * 9 + k + 1);
                }
            }

        dfs();
        for (int i = 1; i <= top; i++)
        {
            auto t = op[ans[i]];
            g[t.x][t.y] = t.z;
        }


        puts(g[0]);
        //puts("");
    }

    return 0;
}

参考链接
https://www.cnblogs.com/ivan-count/p/7355431.html

https://www.cnblogs.com/grenet/p/3145800.html

https://blog.csdn.net/whereisherofrom/article/details/79220897

https://www.luogu.org/blog/ONE-PIECE/qian-tan-dlx 【模板】舞蹈链（DLX）

作者: itdef
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原文地址：https://www.cnblogs.com/itdef/p/14108020.html