HUST 1017 - Exact cover——舞蹈链，精确点覆盖

舞蹈链入门教程：http://www.cnblogs.com/grenet/p/3145800.html

1017 - Exact cover

Time Limit: 15s Memory
Limit: 128MB

Special
Judge Submissions: 7538 Solved: 3868

DESCRIPTION
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.
INPUT
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.
OUTPUT
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".
SAMPLE INPUT

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

SAMPLE OUTPUT

3 2 4 6

HINT

SOURCE
dupeng

舞蹈链模板题

#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];
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;
}
//d为递归深度
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++){
int num,j;
scanf("%d",&num);
while(num--){
scanf("%d",&j);
g.Link(i,j);
}
}
if(!g.Dance(0))printf("NO\n");
else{
printf("%d",g.ansd);
for(int i = 0;i < g.ansd;i++)
printf(" %d",g.ans[i]);
printf("\n");
}
}
return 0;
}