hust1017-Exact cover 舞蹈链之精确覆盖（裸的）

最新推荐文章于 2018-12-07 16:37:32 发布

原创最新推荐文章于 2018-12-07 16:37:32 发布 · 500 阅读

0 ·

CC 4.0 BY-SA版权

舞蹈链专栏收录该内容

6 篇文章

订阅专栏

本文介绍了一种解决精确覆盖问题的方法，通过构建特定的数据结构来高效寻找满足条件的行组合，使得每一列恰好被一行中的1所覆盖。文章提供了一个完整的C++实现代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Sample Output

3 2 4 6

题意：很裸，n行m列，然后给出每行中1的个数和位置，然后问精确覆盖需要几行，并要求打出行数和行号

思路：矩阵已经构造好直接套精确覆盖模板即可是不是非常裸呢^@^

ac代码：

#include <vector>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3fll;
const int maxn=1010;
int L[maxn*100],R[maxn*100],U[maxn*100],D[maxn*100];//节点的上下左右四个方向的链表
int C[maxn*100],H[maxn],cnt[maxn],vis[maxn],ans[maxn],Row[maxn*100];//C列H行cnt列链表中元素个数
int n,m,id,len;
void init(){
    for(int i=0;i<=m;i++){
        cnt[i]=0;U[i]=D[i]=i;
        L[i+1]=i;R[i]=i+1;
    }
    R[m]=0;id=m+1;
    memset(H,-1,sizeof(H));
}
void Link(int r,int c){
    cnt[c]++;C[id]=c;Row[id]=r;
    U[id]=U[c];D[U[c]]=id;
    D[id]=c;U[c]=id;Row[id]=r;
    if(H[r]==-1) H[r]=L[id]=R[id]=id;
    else{
        L[id]=L[H[r]];R[L[H[r]]]=id;
        R[id]=H[r];L[H[r]]=id;
    }
    id++;
}
void Remove(int Size){
    L[R[Size]]=L[Size];
    R[L[Size]]=R[Size];
    for(int i=D[Size];i!=Size;i=D[i]){
        for(int j=R[i];j!=i;j=R[j]){
            U[D[j]]=U[j];D[U[j]]=D[j];
            cnt[C[j]]--;
        }
    }
}
void Resume(int Size){
    for(int i=D[Size];i!=Size;i=D[i]){
        for(int j=R[i];j!=i;j=R[j]){
            U[D[j]]=j;D[U[j]]=j;
            cnt[C[j]]++;
        }
    }
    L[R[Size]]=Size;R[L[Size]]=Size;
}
int Dance(int k){
    int pos,mm=maxn;
    if(R[0]==0){
        len=k;
        return 1;
    }
    for(int i=R[0];i;i=R[i]){
        if(mm>cnt[i]){
            mm=cnt[i];pos=i;
        }
    }
    Remove(pos);
    for(int i=D[pos];i!=pos;i=D[i]){
        ans[k]=Row[i];
        for(int j=R[i];j!=i;j=R[j]) Remove(C[j]);
        if(Dance(k+1)) return 1;
        for(int j=L[i];j!=i;j=L[j]) Resume(C[j]);
    }
    Resume(pos);
    return 0;
}
int main(){
    int u,v,a;
    while(scanf("%d%d",&n,&m)!=-1){
        init();
        for(int i=1;i<=n;i++){
            scanf("%d",&a);
            while(a--){
                scanf("%d",&u);
                Link(i,u);
            }
        }
        int flag=Dance(0);
        if(flag==0) printf("NO\n");
        else{
            printf("%d",len);
            for(int i=0;i<len;i++) printf(" %d",ans[i]);
            printf("\n");
        }
    }
    return 0;
}