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题意:求矩形内最大值。二维RMQ。
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 17 using namespace std; 18 19 #define mem0(a) memset(a, 0, sizeof(a)) 20 #define lson l, m, rt << 1 21 #define rson m + 1, r, rt << 1 | 1 22 #define define_m int m = (l + r) >> 1 23 #define Rep(a, b) for(int a = 0; a < b; a++) 24 #define lowbit(x) ((x) & (-(x))) 25 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 26 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 27 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 28 29 typedef double db; 30 typedef long long LL; 31 typedef pair<int, int> pii; 32 typedef multiset<int> msi; 33 typedef multiset<int>::iterator msii; 34 typedef set<int> si; 35 typedef set<int>::iterator sii; 36 typedef vector<int> vi; 37 38 const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1}; 39 const int dy[8] = {0, -1, 0, 1, -1, 1, 1, -1}; 40 const int maxn = 1e5 + 7; 41 const int maxm = 1e5 + 7; 42 const int maxv = 1e7 + 7; 43 const int MD = 1e9 +7; 44 const int INF = 1e9 + 7; 45 const double PI = acos(-1.0); 46 const double eps = 1e-10; 47 48 int a[300][300], t[301], f[300][9][300][9], n, m; 49 50 void Init_RMQ() { 51 for (int i = 0; i < n; i++) { 52 for (int j = 0; j < m; j++) { 53 f[i][0][j][0] = a[i][j]; 54 } 55 } 56 for (int dx = 0; dx <= 8; dx++) { 57 for (int dy = 0; dy <= 8; dy++) { 58 if (dx == 0 && dy == 0) continue; 59 for (int i = 0; i + (1 << dx) <= n; i++) { 60 for (int j = 0; j + (1 << dy) <= m; j++) { 61 int &p = f[i][dx][j][dy]; 62 if (dx) p = max(f[i][dx - 1][j][dy], f[i + (1 << (dx - 1))][dx - 1][j][dy]); 63 else p = max(f[i][dx][j][dy - 1], f[i][dx][j + (1 << (dy - 1))][dy - 1]); 64 } 65 } 66 } 67 } 68 } 69 70 int RMQ(int x1, int y1, int x2, int y2) { 71 int l1 = t[x2 - x1 + 1], l2 = t[y2 - y1 + 1]; 72 return max(f[x1][l1][y1][l2], max(f[x1][l1][y2 - (1 << l2) + 1][l2], 73 max(f[x2 - (1 << l1) + 1][l1][y1][l2], f[x2 - (1 << l1) + 1][l1][y2 - (1 << l2) + 1][l2]))); 74 } 75 76 int main() { 77 //freopen("in.txt", "r", stdin); 78 for (int i = 1; i <= 300; i++) { 79 int j = 0; 80 while ((1 << (j + 1)) <= i) j++; 81 t[i] = j; 82 } 83 int x1, x2, y1, y2, q; 84 while (cin >> n >> m) { 85 for (int i = 0; i < n; i++) { 86 for (int j = 0; j < m; j++) { 87 scanf("%d", &a[i][j]); 88 } 89 } 90 Init_RMQ(); 91 cin >> q; 92 for (int i = 0; i < q; i++) { 93 scanf("%d%d%d%d", &x1, &y1, &x2, &y2); 94 x1--; x2--; y1--; y2--; 95 int res = RMQ(x1, y1, x2, y2); 96 printf("%d ", res); 97 if (a[x1][y1] == res || a[x2][y1] == res || a[x1][y2] == res || a[x2][y2] == res) puts("yes"); 98 else puts("no"); 99 } 100 } 101 return 0; 102 }
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原文地址:http://www.cnblogs.com/jklongint/p/4418994.html