HUST1017 Exact cover(Dancing Links)Kuangbin模板带解释

Status
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
就是裸Dancing Links熟悉一下

#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];    //U D R L用来记录某个标号的节点的上下左右节点的标号    //Row Col用来记录某个标号的节点在矩阵中的行号和列号      int H[MaxN], S[MaxM];    //H是行头 S用来保存某一列中1的数量    int ansd, ans[MaxN];    void init(int _n,int _m)    {        n = _n;        m = _m;        //初始化列头        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;        }    }    //对某一列进行删除 同时删除列中为1的行    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]];            }    }    //反着恢复状态    void resume(int c)    {        for(int i = U[c];i != c;i = U[i])            for(int j = L[i];j != i;j = L[j])            {                U[D[j]]=D[U[j]]=j;                ++S[Col[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];        //一个优化  找到列中包含1最多的列 因为这样有助于减少递归深度 （很显然1多了 删掉的行也多 矩阵缩小得就快）        for(int i = R[0];i != 0;i = R[i])            if(S[i] < S[c])                c = i;        remove(c);        //搜索        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(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    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;}