HUST 1017 Exact cover （精确覆盖|Dancing Links模板题）

最新推荐文章于 2019-07-11 16:57:22 发布

godspeedkaka

最新推荐文章于 2019-07-11 16:57:22 发布

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CC 4.0 BY-SA版权

文章标签： HUST 精确覆盖 Dancing Links

本文链接：https://blog.youkuaiyun.com/u014664226/article/details/50621208

本文探讨了使用DLX算法解决精确覆盖问题的方法，详细解释了算法的实现步骤，并通过实例展示了如何应用该算法来求解具体问题。

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题意：精确覆盖问题。

思路：测试模板。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define LL long long
#define pii pair<int, int>
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

//const int MAXN = 5000000 + 5;
//const int INF = 0x3f3f3f3f;
const int maxnode = 110000;
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];//H是每一行的头指针，S是每一列的节点数量 
    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;


int main() {
    //freopen("input.txt", "r", stdin);
	int n, m;
	while(cin >> n >> m) {
		dlx.init(n, m);
		for(int i = 1; i <= n; i++) {
			int cnt;
			scanf("%d", &cnt);
			for(int j = 1; j <= cnt; j++) {
				int tmp;
				scanf("%d", &tmp);
				dlx.Link(i, tmp);
			}
		}
		bool ok = dlx.Dance(0);
		if(!ok) puts("NO");
		else {
			printf("%d", dlx.ansd);
			for(int i = 0; i < dlx.ansd; i++) printf(" %d", dlx.ans[i]);
			puts("");
		}
	}
    return 0;
}