POJ 3740 Easy Finding 跳舞链模板

题目：

题意:

思路：

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

using namespace std;
//X是点数，N是行数，M是列数
const int X = 100000 + 10, N = 20 + 10, M = 300 + 10, INF = 0x3f3f3f3f;

struct DLX
{
int U[X], D[X], L[X], R[X], row[X], col[X];
int H
, S[M];
int head, sz, tot, n, m, ans
;
void init(int _n, int _m)
{
n = _n, m = _m;
for(int i = 0; i <= m; i++) //初始化列头
L[i] = i-1, R[i] = i+1, U[i] = D[i] = i, S[i] = 0;
head = 0, tot = 0, sz = m;
L[head] = m, R[m] = head;
for(int i = 1; i <= n; i++) H[i] = -1; //初始化行头
}
void link(int r, int c) /*插入一个新点，注意插入的位置和此点在原矩阵中的位置是上下左右都颠倒的，
但是通过row和col数组记录了点在原矩阵中的位置，所以没有关系*/
{
++S[col[++sz]=c];
row[sz] = r;
D[sz] = D[c], U[D[c]] = sz;
U[sz] = c, D[c] = sz;
if(H[r] < 0) H[r] = L[sz] = R[sz] = sz;
else R[sz] = R[H[r]], L[R[H[r]]] = sz, L[sz] = H[r], R[H[r]] = sz;
}
void del(int c) //删除在某一列出现1元素的所有行
{
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])
D[U[j]] = D[j], U[D[j]] = U[j], --S[col[j]];
}
void recover(int c) //删除的逆过程，注意次序和删除正好一反，但其实次序无所谓，和删除一样也可以
{
for(int i = U[c]; i != c; i = U[i])
for(int j = L[i]; j != i; j = L[j])
D[U[j]] = U[D[j]] = j, ++S[col[j]];
R[L[c]] = L[R[c]] = c;
}
bool dance(int dep) //核心部分，其实就是dfs
{
if(R[head] == head)
{
tot = dep-1; return true;
}
int c = R[head];
for(int i = R[head]; i != head; i = R[i])//选取出现1元素次数最少的一列进行删除
if(S[i] < S[c]) c = i;
del(c);
for(int i = D[c]; i != c; i = D[i])
{
ans[dep] = row[i];
for(int j = R[i]; j != i; j = R[j]) del(col[j]);
if(dance(dep + 1)) return true;
for(int j = L[i]; j != i; j = L[j]) recover(col[j]);
}
recover(c);
return false;
}
}dlx;
int main()
{
int n, m;
while(~ scanf("%d%d", &n, &m))
{
dlx.init(n, m);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
{
int val;
scanf("%d", &val);
if(val == 1) dlx.link(i, j);
}
puts(dlx.dance(1) ? "Yes, I found it" : "It is impossible");
}
return 0;
}