ZOJ 3209 Treasure Map (DLX精確覆蓋問題)

題目大意：

給出一個n*m的矩形, n, m <= 30, 從p <= 500個矩形中選擇一些矩形使得這些矩形不重合但剛好拼湊出n*m的這個矩形(矩形位置都不能移動), 求從給出的矩形中最少需要挑出幾個才能滿足這個條件

大致思路：

就是將n*m個小的1*1的正方形視作一個單位做精確覆蓋問題就好了

最壞情況下900列, 500行, 直接用DLX就可以

代碼如下：

Result  :  Accepted     Memory  :  10828 KB     Time  :  70 ms

/*
 * Author: Gatevin
 * Created Time:  2015/10/4 18:10:46
 * File Name: Sakura_Chiyo.cpp
 */
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;

#define maxnode 450010
#define maxn 510
#define maxm 1000

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;
    }
    void Dance(int dep)
    {
        if(dep >= ansd) return;
        if(R[0] == 0)
        {
            ansd = min(ansd, dep);
            return;
        }
        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[dep] = Row[i];
            for(int j = R[i]; j != i; j = R[j]) remove(Col[j]);
            Dance(dep + 1);
            for(int j = L[i]; j != i; j = L[j]) resume(Col[j]);
        }
        resume(c);
        return;
    }
    void solve()
    {
        int N, M, p;
        scanf("%d %d %d", &N, &M, &p);
        init(p, N*M);
        for(int i = 1; i <= p; i++)
        {
            int x1, y1, x2, y2;
            scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
            for(int x = x1; x < x2; x++)
                for(int y = y1; y < y2; y++)
                    Link(i, x*M + y + 1);
        }
        ansd = p + 1;
        Dance(0);
        if(ansd == p + 1)
            puts("-1");
        else printf("%d\n", ansd);
        return;
    }
};

DLX dlx;

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
        dlx.solve();
    return 0;
}