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-;
 const int mod = 1e9+;
 const int inf = ;
 const int dir[][] = { {-, }, {, }, {, -}, {, } };
 const int maxn = ;
 const int maxNode = ;
 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[][];
     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 = ;
         int vis[];
         mem(vis);
         for(int i = R[]; i!=; i = R[i]) {
             if(!vis[i]) {
                 cnt++;
                 vis[i] = ;
                 for(int j = D[i]; j!=i; j = D[j]) {
                     for(int k = R[j]; k!=j; k = R[k]) {
                         vis[col[k]] = ;
                     }
                 }
             }
         }
         return cnt;
     }
     void dfs(int d) {
         if(d+h()>=deep)
             return ;
         if(R[] == ) {
             deep = min(deep, d);
             return ;
         }
         int c = R[];
         for(int i = R[]; i!=; 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+);
             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 = ; i<=n; i++) {
             R[i] = i+;
             L[i] = i-;
             U[i] = i;
             D[i] = i;
         }
         deep = inf;
         mem(S);
         R[n] = ;
         L[] = n;
         sz = n;
     }
     void solve() {
         mem(g);
         int cnt = , x;
         for(int i = ; i<n; i++) {
             for(int j = ; j<m; j++) {
                 scanf("%d", &x);
                 if(x)
                     g[i][j] = ++cnt;
                 else
                     g[i][j] = ;
             }
         }
         scanf("%d%d", &n1, &m1);
         int r = , tmp = n;
         n = cnt;
         init();
         for(int i = ; i<tmp-n1+; i++) {
             for(int j = ; j<m-m1+; 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();
         printf("%d\n", deep);
     }
 }dlx;
 int main()
 {
     while(scanf("%d%d", &dlx.n, &dlx.m)!=EOF) {
         dlx.solve();
     }
     return ;
 }