HDU2888 Check Corners（二维RMQ）
有一个矩阵，每次查询一个子矩阵，判断这个子矩阵的最大值是不是在这个子矩阵的四个角上
裸的二维RMQ
1 #pragma comment(linker, "/STACK:1677721600") 2 #include <map> 3 #include <set> 4 #include <stack> 5 #include <queue> 6 #include <cmath> 7 #include <ctime> 8 #include <vector> 9 #include <cstdio> 10 #include <cctype> 11 #include <cstring> 12 #include <cstdlib> 13 #include <iostream> 14 #include <algorithm> 15 using namespace std; 16 #define INF 0x3f3f3f3f 17 #define inf (-((LL)1<<40)) 18 #define lson k<<1, L, (L + R)>>1 19 #define rson k<<1|1, ((L + R)>>1) + 1, R 20 #define mem0(a) memset(a,0,sizeof(a)) 21 #define mem1(a) memset(a,-1,sizeof(a)) 22 #define mem(a, b) memset(a, b, sizeof(a)) 23 #define FIN freopen("in.txt", "r", stdin) 24 #define FOUT freopen("out.txt", "w", stdout) 25 #define rep(i, a, b) for(int i = a; i <= b; i ++) 26 #define dec(i, a, b) for(int i = a; i >= b; i --) 27 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 32 33 //typedef __int64 LL; 34 typedef long long LL; 35 const int MAXN = 50000 + 100; 36 const int MAXM = 110000; 37 const double eps = 1e-8; 38 LL MOD = 1000000007; 39 40 int m, n; 41 int mx[302][9][302][9]; 42 int idx[302], q, lx, ly, rx, ry; 43 44 void rmq_init(int m, int n) { 45 for(int i = 0; (1<<i) <= m; i ++) { 46 for(int j = 0; (1<<j) <= n; j ++) { 47 if(i == 0 && j == 0) continue; 48 int len2 = (1 << j), len1 = (1 << i); 49 for(int x = 1; x + len1 - 1 <= m; x ++) { 50 for(int y = 1; y + len2 - 1 <= n; y ++) { 51 if(i == 0) mx[x][i][y][j] = max(mx[x][i][y][j - 1], mx[x][i][y + (len2 >> 1)][j - 1]); 52 else mx[x][i][y][j] = max(mx[x][i - 1][y][j], mx[x + (len1 >> 1)][i - 1][y][j]); 53 } 54 } 55 } 56 } 57 for(int i = 1; i <= m || i <= n; i ++) { 58 idx[i] = 0; 59 while((1 << (idx[i] + 1)) <= i) idx[i] ++; 60 } 61 } 62 63 int rmq(int lx, int rx, int ly, int ry) { 64 int a = idx[rx - lx + 1], la = (1 << a); 65 int b = idx[ry - ly + 1], lb = (1 << b); 66 return max(max(max(mx[lx][a][ly][b], 67 mx[rx - la + 1][a][ly][b]), 68 mx[lx][a][ry - lb + 1][b]), 69 mx[rx - la + 1][a][ry - lb + 1][b]); 70 } 71 72 int main() 73 { 74 //FIN; 75 while(~scanf("%d %d", &m, &n)) { 76 mem0(mx); 77 rep (i, 1, m) rep (j, 1, n) 78 scanf("%d", &mx[i][0][j][0]); 79 rmq_init(m, n); 80 scanf("%d", &q); 81 while(q--) { 82 scanf("%d %d %d %d", &lx, &ly, &rx, &ry); 83 int ma = rmq(lx, rx, ly, ry); 84 printf("%d %s\n", ma, ma == mx[lx][0][ly][0] || ma == mx[lx][0][ry][0] 85 || ma == mx[rx][0][ly][0] || ma == mx[rx][0][ry][0] 86 ? "yes" : "no"); 87 } 88 } 89 return 0; 90 }
