HUST 1017 - Exact cover——舞蹈链，精确点覆盖

原创于 2017-07-21 11:37:21 发布 · 411 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#算法

算法同时被 2 个专栏收录

246 篇文章

订阅专栏

舞蹈链

2 篇文章

订阅专栏

本文介绍了一种使用舞蹈链算法解决精确覆盖问题的方法。通过构建特定矩阵并利用舞蹈链进行搜索，可以找出使得每列恰好有一个1的行选择方案。文章提供了完整的C++实现代码及示例。

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

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;
}