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codeforces 713D D. Animals and Puzzle 二分+二维rmq

题目链接

 

给一个01矩阵, 然后每个询问给出两个坐标(x1, y1), (x2, y2)。 问你这个范围内的最大全1正方形的边长是多少。

 

我们dp算出以i, j为右下角的正方形边长最大值。 然后用二维st表预处理出所有的最大值。 对于每个询问, 我们二分一个值mid, 查询(x1 + mid -1, y1 + mid -1), (x2, y2)这个范围内的最大值是否大于mid 。如果大于的话就说明在(x1, y1), (x2, y2)范围内存在一个边长为mid的正方形。

#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, n, a) for(int i = a; i<n; i++)#define fi first#define se secondtypedef complex <double> cmx;typedef pair<int, int> pll;const double PI = acos(-1.0);const double eps = 1e-8;const int mod = 1e9+7;const int inf = 1061109567;const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };int dp[1001][1001][11][11], mm[1005];void initRmq(int n, int m){    mm[0] = -1;    for(int i = 1; i <= max(n, m); i++) {        mm[i] = ((i&(i-1)) == 0) ? mm[i-1] + 1 : mm[i-1];    }    for (int ii = 0; ii <= mm[n]; ii ++) {        for (int jj = 0; jj <= mm[m]; jj ++) {            if (ii + jj) {                for (int i = 1; i + (1<<ii) - 1 <= n; i ++) {                    for(int j = 1; j + (1<<jj) - 1 <= m; j ++) {                        if (ii) {                            dp[i][j][ii][jj] = max(dp[i][j][ii-1][jj], dp[i+(1<<(ii-1))][j][ii-1][jj]);                        } else {                            dp[i][j][ii][jj] = max(dp[i][j][ii][jj-1], dp[i][j+(1<<(jj-1))][ii][jj-1]);                        }                    }                }            }        }    }}int rmq(int x1, int y1, int x2, int y2){    int k1 = mm[x2-x1+1];    int k2 = mm[y2-y1+1];    x2 = x2 - (1<<k1) + 1;    y2 = y2 - (1<<k2) + 1;    return max(max(dp[x1][y1][k1][k2], dp[x1][y2][k1][k2]), max((dp[x2][y1][k1][k2]), dp[x2][y2][k1][k2]));}int main(){    int n, m, x, q, x1, y1, x2, y2;    cin>>n>>m;    for (int i = 1; i <= n; i++) {        for (int j = 1; j <= m; j++) {            scanf("%d", &x);            if (x)                dp[i][j][0][0] = min(min(dp[i-1][j][0][0], dp[i][j-1][0][0]), dp[i-1][j-1][0][0]) + 1;        }    }    initRmq(n, m);    cin>>q;    while (q--) {        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);        int l = 0, r = min(x2 - x1, y2 - y1) + 1;        int ans;        while (l <= r) {            int mid = l + r >> 1;            if (rmq(x1 + mid - 1, y1 + mid - 1, x2, y2) >= mid) {                ans = mid;                l = mid + 1;            } else {                r = mid - 1;            }        }        printf("%d\n", ans);    }    return 0;}

 

codeforces 713D D. Animals and Puzzle 二分+二维rmq