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 second
typedef 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;
}

 

posted on 2016-10-04 10:31  yohaha  阅读(107)  评论(0编辑  收藏

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