AcWing 1067. 精确覆盖问题 DLX
作者：互联网
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 << "=================================================>\n";
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
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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 }
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标签：right,NULL,1067,AcWing,table,DLX,col,row 来源： https://www.cnblogs.com/itdef/p/14108020.html