hust 1017 - Exact cover (DLX)

1017 - Exact cover

时间限制：15秒 内存限制：128兆

自定评测 6414
次提交 3374 次通过

题目描述
There is an N*M matrix with only 0s and 1s, (1 <= N,M <= 1000). An exact cover is a selection of rows such that every column has a 1 in exactly one of the selected rows. Try to find out the selected rows.
输入
There are multiply test cases. First line: two integers N, M; The following N lines: Every line first comes an integer C(1 <= C <= 100), represents the number of 1s in this row, then comes C integers: the index of the columns whose value is 1 in this row.
输出
First output the number of rows in the selection, then output the index of the selected rows. If there are multiply selections, you should just output any of them. If there are no selection, just output "NO".
样例输入

6 7
3 1 4 7
2 1 4
3 4 5 7
3 3 5 6
4 2 3 6 7
2 2 7

样例输出

3 2 4 6

复习一下dancing link

/*======================================================
# Author: whai
# Last modified: 2015-10-08 15:25
# Filename: hust1017_v1.cpp
======================================================*/
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <set>
#include <map>

using namespace std;

#define LL __int64
#define PB push_back
#define P pair<int, int>
#define X first
#define Y second

const int MAXN = 1010;
const int MAXM = 1010;
const int N = MAXN * MAXM;
struct DLX {
int n, m, size;
int U
, D
, R
, L
, row
, col
;
int H[MAXN], S[MAXM];
int ans[MAXN], ans_cnt;	//ans_cnt表示覆盖所需要的数量，ans保存覆盖所需要的部分
void init(int _n, int _m) {
ans_cnt = -1;
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;
}
void dance(int d) {
//剪枝下
//if(ans_cnt != -1 && ans_cnt <= d) return;	//改，根据题目而定，如果要求用最小的东西覆盖，则把这个加上
if (R[0] == 0) {
//if(ans_cnt == -1) ans_cnt = d; //改，根据题目而定，如果要求用最小的东西覆盖，则把这个加上
//else if(d < ans_cnt) ans_cnt = d; //改，根据题目而定，如果要求用最小的东西覆盖，则把这个加上
ans_cnt = d; //改，根据题目而定，如果要求用最小的东西覆盖，则把这个去掉
return;
}
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]);
dance(d + 1);
if (ans_cnt != -1) return ;	//改，根据题目而定，如果要求用最小的东西覆盖，则把这个去掉
for (int j = L[i]; j != i; j = L[j]) resume(col[j]);
}
resume(c);
}
}dlx;

int main() {
int n, m;
while(scanf("%d%d", &n, &m) != EOF) {
dlx.init(n, m);
for(int i = 1; i <= n; ++i) {
int c, j;
scanf("%d", &c);
while(c--) {
scanf("%d", &j);
dlx.link(i, j);
}
}
dlx.dance(0);
if(dlx.ans_cnt == -1) puts("NO");
else {
printf("%d", dlx.ans_cnt);
for(int i = 0; i < dlx.ans_cnt; ++i) {
printf(" %d", dlx.ans[i]);
}
puts("");
}
}
return 0;
}