HUST 1017

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 73 1 4 72 1 43 4 5 73 3 5 64 2 3 6 72 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;}