FZU Problem 1686 神龙的难题重复覆盖

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。
  1 #include <iostream>
  2 #include <vector>
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <algorithm>
  6 #include <cmath>
  7 #include <map>
  8 #include <set>
  9 #include <string>
 10 #include <queue>
 11 using namespace std;
 12 #define pb(x) push_back(x)
 13 #define ll long long
 14 #define mk(x, y) make_pair(x, y)
 15 #define lson l, m, rt<<1
 16 #define mem(a) memset(a, 0, sizeof(a))
 17 #define rson m+1, r, rt<<1|1
 18 #define mem1(a) memset(a, -1, sizeof(a))
 19 #define mem2(a) memset(a, 0x3f, sizeof(a))
 20 #define rep(i, a, n) for(int i = a; i<n; i++)
 21 #define ull unsigned long long
 22 typedef pair<int, int> pll;
 23 const double PI = acos(-1.0);
 24 const double eps = 1e-8;
 25 const int mod = 1e9+7;
 26 const int inf = 1061109567;
 27 const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
 28 const int maxn = 305;
 29 const int maxNode = 5000;
 30 struct DLX {
 31     int L[maxNode], R[maxNode], U[maxNode], D[maxNode], row[maxNode], col[maxNode];
 32     int S[maxn], H[maxn], deep, ans[maxn], sz, n, m, k, n1, m1;
 33     int g[20][20];
 34     void remove(int c) {
 35         for(int i = D[c]; i!=c; i = D[i]) {
 36             L[R[i]] = L[i];
 37             R[L[i]] = R[i];
 38         }
 39     }
 40     void resume(int c) {
 41         for(int i = U[c]; i!=c; i = U[i]) {
 42             L[R[i]] = i;
 43             R[L[i]] = i;
 44         }
 45     }
 46     int h() {
 47         int cnt = 0;
 48         int vis[250];
 49         mem(vis);
 50         for(int i = R[0]; i!=0; i = R[i]) {
 51             if(!vis[i]) {
 52                 cnt++;
 53                 vis[i] = 1;
 54                 for(int j = D[i]; j!=i; j = D[j]) {
 55                     for(int k = R[j]; k!=j; k = R[k]) {
 56                         vis[col[k]] = 1;
 57                     }
 58                 }
 59             }
 60         }
 61         return cnt;
 62     }
 63     void dfs(int d) {
 64         if(d+h()>=deep)
 65             return ;
 66         if(R[0] == 0) {
 67             deep = min(deep, d);
 68             return ;
 69         }
 70         int c = R[0];
 71         for(int i = R[0]; i!=0; i = R[i])
 72             if(S[c]>S[i])
 73                 c = i;
 74         for(int i = D[c]; i!=c; i = D[i]) {
 75             remove(i);
 76             for(int j = R[i]; j!=i; j = R[j])
 77                 remove(j);
 78             dfs(d+1);
 79             for(int j = L[i]; j!=i; j = L[j])
 80                 resume(j);
 81             resume(i);
 82         }
 83         return ;
 84     }
 85     void add(int r, int c) {
 86         sz++;
 87         row[sz] = r;
 88         col[sz] = c;
 89         S[c]++;
 90         U[sz] = U[c];
 91         D[sz] = c;
 92         D[U[c]] = sz;
 93         U[c] = sz;
 94         if(~H[r]) {
 95             R[sz] = H[r];
 96             L[sz] = L[H[r]];
 97             L[R[sz]] = sz;
 98             R[L[sz]] = sz;
 99         } else {
100             H[r] = L[sz] = R[sz] = sz;
101         }
102     }
103     void init(){
104         mem1(H);
105         for(int i = 0; i<=n; i++) {
106             R[i] = i+1;
107             L[i] = i-1;
108             U[i] = i;
109             D[i] = i;
110         }
111         deep = inf;
112         mem(S);
113         R[n] = 0;
114         L[0] = n;
115         sz = n;
116     }
117     void solve() {
118         mem(g);
119         int cnt = 0, x;
120         for(int i = 0; i<n; i++) {
121             for(int j = 0; j<m; j++) {
122                 scanf("%d", &x);
123                 if(x)
124                     g[i][j] = ++cnt;
125                 else
126                     g[i][j] = 0;
127             }
128         }
129         scanf("%d%d", &n1, &m1);
130         int r = 0, tmp = n;
131         n = cnt;
132         init();
133         for(int i = 0; i<tmp-n1+1; i++) {
134             for(int j = 0; j<m-m1+1; j++) {
135                 r++;
136                 for(int k1 = i; k1<min(i+n1, tmp); k1++) {
137                     for(int k2 = j; k2<min(j+m1, m); k2++) {
138                         if(g[k1][k2]) {
139                             add(r, g[k1][k2]);
140                         }
141                     }
142                 }
143             }
144         }
145         dfs(0);
146         printf("%d
", deep);
147     }
148 }dlx;
149 int main()
150 {
151     while(scanf("%d%d", &dlx.n, &dlx.m)!=EOF) {
152         dlx.solve();
153     }
154     return 0;
155 }
FZU Problem 1686 神龙的难题 重复覆盖

FZU Problem 1686 神龙的难题重复覆盖
