标签:ems == signed names base name inf ons for
就是板题。。
查询子矩阵中最大的元素。。。然后看看是不是四个角落的 是就是yes 不是就是no 判断一下就好了
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 301, INF = 0x7fffffff; int n, m; int dp[maxn][maxn][9][9], a[maxn][maxn]; int rmq(int x1, int y1, int x2, int y2) { int kx = 0, ky = 0; while ((1 << (1 + kx)) <= x2 - x1 + 1) kx++; while ((1 << (1 + ky)) <= y2 - y1 + 1) ky++; int m1 = dp[x1][y1][kx][ky]; int m2 = dp[x2 - (1 << kx) + 1][y1][kx][ky]; int m3 = dp[x1][y2 - (1 << ky) + 1][kx][ky]; int m4 = dp[x2 - (1 << kx) + 1][y2 - (1 << ky) + 1][kx][ky]; return max(max(m1, m2), max(m3, m4)); } int main() { while(cin>> n >> m) { rap(i, 1, n) rap(j, 1, m) { scanf("%d", &a[i][j]); dp[i][j][0][0] = a[i][j]; } for (int i = 0; (1 << i) <= n; i++) { for (int j = 0; (1 << j) <= m; j++) { if (i == 0 && j == 0) continue; for (int row = 1; row + (1 << i) - 1 <= n; row++) for (int col = 1; col + (1 << j) - 1 <= m; col++) { //当x或y等于0的时候,就相当于一维的RMQ了 //if(i == 0) dp[row][col][i][j] = max(dp[row][col][i][j - 1], dp[row][col + (1 << (j - 1))][i][j - 1]); if (j == 0) dp[row][col][i][j] = max(dp[row][col][i - 1][j], dp[row + (1 << (i - 1))][col][i - 1][j]); else dp[row][col][i][j] = max(dp[row][col][i][j - 1], dp[row][col + (1 << (j - 1))][i][j - 1]); } } } int q; scanf("%d", &q); while(q--) { int x1, y1, x2, y2; scanf("%d%d%d%d", &x1, &y1, &x2, &y2); int h = rmq(x1, y1, x2, y2); int flag = 0; if(a[x1][y1] == h || a[x2][y2] == h || a[x1][y2] == h || a[x2][y1] == h) flag = 1; printf("%d %s\n", h, flag?"yes":"no"); } } return 0; }
Check Corners HDU - 2888(二维RMQ)
标签:ems == signed names base name inf ons for
原文地址:https://www.cnblogs.com/WTSRUVF/p/9450927.html