ZOJ 3209 Treasure Map 精确覆盖

题目链接

精确覆盖的模板题， 把每一个格子当成一列就可以。

S忘记初始化TLE N次， 哭晕在厕所......

 #include<bits/stdc++.h>
 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, sz, n, m, num;
     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[j]]++;
                 D[U[j]] = j;
                 U[D[j]] = j;
             }
         R[L[c]] = c;
         L[R[c]] = c;
     }
     void dfs(int d) {
         if(d>=deep)
             return ;
         if(R[] == ) {
             deep = min(d, deep);
             return ;
         }
         int c = R[];
         for(int i = R[]; i!=; i = R[i])
             if(S[c]>S[i])
                 c = i;
         remove(c);
         for(int i = D[c]; i!=c; i = D[i]) {
             for(int j = R[i]; j!=i; j = R[j])
                 remove(col[j]);
             dfs(d+);
             for(int j = L[i]; j!=i; j = L[j])
                 resume(col[j]);
         }
         resume(c);
         return ;
     }
     void add(int r, int c) {
         sz++;
         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-;
             D[i] = U[i] = i;
         }
         R[n] = ;
         L[] = n;
         mem(S);
         sz = n;
         deep = inf;
     }
     void solve() {
         scanf("%d%d%d", &n, &m, &num);
         n *= m;
         init();
         int x1, y1, x2, y2;
         int cnt = ;
         while(num--) {
             scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
             for(int i = x1+; i<=x2; i++) {
                 for(int j = y1+; j<=y2; j++) {
                     add(cnt, (i-)*m+j);
                 }
             }
             cnt++;
         }
         dfs();
         if(deep == inf) {
             puts("-1");
         } else {
             printf("%d\n", deep);
         }
     }
 }dlx;
 int main()
 {
     int t;
     cin>>t;
     while(t--) {
         dlx.solve();
     }
 }