FZU 1686 神龙的难题（重复覆盖|Dancing Links）

题意：这是个剑与魔法的世界.英雄和魔物同在,动荡和安定并存.但总的来说,库尔特王国是个安宁的国家,人民安居乐业,魔物也比较少.但是.总有一些魔物不时会进入城市附近,干扰人民的生活.就要有一些人出来守护居民们不被魔物侵害.魔法使艾米莉就是这样的一个人.她骑着她的坐骑,神龙米格拉一起消灭干扰人类生存的魔物,维护王国的安定.艾米莉希望能够在损伤最小的前提下完成任务.每次战斗前,她都用时间停止魔法停住时间,然后米格拉他就可以发出火球烧死敌人.米格拉想知道,他如何以最快的速度消灭敌人,减轻艾米莉的负担.

思路：将每一一个1看成一列，将每一个可以攻击的子矩阵看成一行，然后用DLX重复覆盖来解决即可。

#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 = 100000;
const int MaxM = 250;
const int MaxN = 250;
int G[20][20], id[20][20];
struct DLX
{
    int n,m,size;
    int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];
    int H[MaxN],S[MaxN];
    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;
        ansd = INF;
    }
    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)
    {
        for(int i = D[c];i != c;i = D[i])
            L[R[i]] = L[i], R[L[i]] = R[i];
    }
    void resume(int c)
    {
        for(int i = U[c];i != c;i = U[i])
            L[R[i]]=R[L[i]]=i;
    }
    bool v[maxnode];
    int f()
    {
        int ret = 0;
        for(int c = R[0];c != 0;c = R[c])v[c] = true;
        for(int c = R[0];c != 0;c = R[c])
            if(v[c])
            {
                ret++;
                v[c] = false;
                for(int i = D[c];i != c;i = D[i])
                    for(int j = R[i];j != i;j = R[j])
                        v[Col[j]] = false;
            }
        return ret;

    }
    void Dance(int d)
    {
        if(d + f() >= ansd) return;
		if(R[0] == 0) {
        	ansd = min(ansd, d);
        	return;
		}
        int c = R[0];
        for(int i = R[0];i != 0;i = R[i])
            if(S[i] < S[c])
                c = i;
        for(int i = D[c];i != c;i = D[i])
        {
            remove(i);
            for(int j = R[i];j != i;j = R[j])remove(j);
            Dance(d+1);
            for(int j = L[i];j != i;j = L[j])resume(j);
            resume(i);
        }
    }
} dlx;
int main() {
    //freopen("input.txt", "r", stdin);
    int n, m;
	while(cin >> n >> m) {
		int cnt = 0;
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= m; j++) {
				scanf("%d", &G[i][j]);
				if(G[i][j]) {
					id[i][j] = ++cnt;
				}
			}
		}
		int n1, m1;
		scanf("%d%d", &n1, &m1);
		dlx.init((n-n1+1)*(m-m1+1), cnt);
		for(int i = 1; i <= n-n1+1; i++) {
			for(int j = 1; j <= m-m1+1; j++) {
				for(int k = i; k < i+n1; k++) {
					for(int p = j; p < j+m1; p++) {
						if(G[k][p]) dlx.Link((i-1)*(m-m1+1)+j, id[k][p]);
					}
				}
			}
		}
		dlx.Dance(0);
		cout << dlx.ansd << endl;
	}
    return 0;
}