zoukankan      html  css  js  c++  java
  • [HUST 1017] Exact cover

    Exact cover

    Time Limit: 15s Memory Limit: 128MB

    Special Judge Submissions: 6012 Solved: 3185
    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
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    using namespace std;
    #define MaxNode 100010
    #define MaxN 1010
    #define MaxM 1010
    
    struct DLX
    {
        int n,m,size;
        int U[MaxNode],D[MaxNode],R[MaxNode],L[MaxNode];
        int 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;
            U[size]=U[c];
            D[U[c]]=size;
            D[size]=c;
            U[c]=size;
            if(H[r]==-1) H[r]=L[size]=R[size]=size;
            else
            {
                L[size]=L[H[r]];
                R[L[H[r]]]=size;
                R[size]=H[r];
                L[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])
                {
                    U[D[j]]=j;
                    D[U[j]]=j;
                    S[Col[j]]++;
                }
            }
            L[R[c]] =c;
            R[L[c]] =c;
        }
        bool Dance(int d)
        {
            if(R[0]==0)
            {
                ansd=d;
                return 1;
            }
            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]);  //移除
                if(Dance(d+1)) return 1;
                for(int j=L[i];j!=i;j=L[j]) Resume(Col[j]);  //回标
            }
            Resume(c);
            return 0;
        }
    }g;
    int main()
    {
        int n,m;
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            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
    ");
            else
            {
                printf("%d",g.ansd);
                for(int i=0;i<g.ansd;i++)
                    printf(" %d",g.ans[i]);
                printf("
    ");
            }
        }
        return 0;
    }
    趁着还有梦想、将AC进行到底~~~by 452181625
  • 相关阅读:
    CouchDB
    在 Fabric 中使用私有数据
    Hyperledger Fabric 踩坑汇总
    书单
    HyperLedger Fabric 资料网址大全
    Hyperledger composer
    Hyperledger Fabric
    [转]以太坊智能合约编程之菜鸟教程
    [转]Ethereum-智能合约最佳实践
    [转]工作量证明(PoW)权益证明(PoS)和委任权益证明(DPoS)区别
  • 原文地址:https://www.cnblogs.com/hate13/p/4183395.html
Copyright © 2011-2022 走看看