FZU Problem 1686 神龙的难题 重复覆盖

题目链接

给出大矩形的长宽， 矩形里面有1,0两个值， 给出小矩形的长宽， 求用最少的小矩形覆盖所有的1.

重复覆盖的模板题。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, a, n) for(int i = a; i<n; i++)
#define ull unsigned long long
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 305;
const int maxNode = 5000;
struct DLX {
    int L[maxNode], R[maxNode], U[maxNode], D[maxNode], row[maxNode], col[maxNode];
    int S[maxn], H[maxn], deep, ans[maxn], sz, n, m, k, n1, m1;
    int g[20][20];
    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]] = i;
            R[L[i]] = i;
        }
    }
    int h() {
        int cnt = 0;
        int vis[250];
        mem(vis);
        for(int i = R[0]; i!=0; i = R[i]) {
            if(!vis[i]) {
                cnt++;
                vis[i] = 1;
                for(int j = D[i]; j!=i; j = D[j]) {
                    for(int k = R[j]; k!=j; k = R[k]) {
                        vis[col[k]] = 1;
                    }
                }
            }
        }
        return cnt;
    }
    void dfs(int d) {
        if(d+h()>=deep)
            return ;
        if(R[0] == 0) {
            deep = min(deep, d);
            return ;
        }
        int c = R[0];
        for(int i = R[0]; i!=0; i = R[i])
            if(S[c]>S[i])
                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);
            dfs(d+1);
            for(int j = L[i]; j!=i; j = L[j])
                resume(j);
            resume(i);
        }
        return ;
    }
    void add(int r, int c) {
        sz++;
        row[sz] = r;
        col[sz] = c;
        S[c]++;
        U[sz] = U[c];
        D[sz] = c;
        D[U[c]] = sz;
        U[c] = sz;
        if(~H[r]) {
            R[sz] = H[r];
            L[sz] = L[H[r]];
            L[R[sz]] = sz;
            R[L[sz]] = sz;
        } else {
            H[r] = L[sz] = R[sz] = sz;
        }
    }
    void init(){
        mem1(H);
        for(int i = 0; i<=n; i++) {
            R[i] = i+1;
            L[i] = i-1;
            U[i] = i;
            D[i] = i;
        }
        deep = inf;
        mem(S);
        R[n] = 0;
        L[0] = n;
        sz = n;
    }
    void solve() {
        mem(g);
        int cnt = 0, x;
        for(int i = 0; i<n; i++) {
            for(int j = 0; j<m; j++) {
                scanf("%d", &x);
                if(x)
                    g[i][j] = ++cnt;
                else
                    g[i][j] = 0;
            }
        }
        scanf("%d%d", &n1, &m1);
        int r = 0, tmp = n;
        n = cnt;
        init();
        for(int i = 0; i<tmp-n1+1; i++) {
            for(int j = 0; j<m-m1+1; j++) {
                r++;
                for(int k1 = i; k1<min(i+n1, tmp); k1++) {
                    for(int k2 = j; k2<min(j+m1, m); k2++) {
                        if(g[k1][k2]) {
                            add(r, g[k1][k2]);
                        }
                    }
                }
            }
        }
        dfs(0);
        printf("%d
", deep);
    }
}dlx;
int main()
{
    while(scanf("%d%d", &dlx.n, &dlx.m)!=EOF) {
        dlx.solve();
    }
    return 0;
}