ZOJ-3955-Saddle Point【17th浙大校赛】

Saddle Point

Chiaki has an n × m matrix A. Rows are numbered from 1 to n from top to bottom and columns are numbered from 1 to m from left to right. The element in the i-th row and the j-th column is Ai, j.

Let M({i1, i2, …, is}, {j1, j2, …, jt}) be the matrix that results from deleting row i1, i2, …, is and column j1, j2, …, jt of A and f({i1, i2, …, is}, {j1, j2, …, jt}) be the number of saddle points in matrix M({i1, i2, …, is}, {j1, j2, …, jt}).

Chiaki would like to find all the value of f({i1, i2, …, is}, {j1, j2, …, jt}). As the output may be very large ((2n - 1)(2m - 1) matrix in total), she is only interested in the value

Note that a saddle point of a matrix is an element which is both the only largest element in its column and the only smallest element in its row.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains four integers n and m (1 ≤ n, m ≤ 1000) – the number of rows and the number of columns.

Each of the next n lines contains m integer Ai, 1, Ai, 2, …, Ai, m (1 ≤ Ai, j ≤ 106), where Ai, j is the integer in the i-th row and the j-th column.

It is guaranteed that neither the sum of all n nor the sum of all m exceeds 5000.

Output

For each test case, output an integer denoting the answer.

Sample Input
2
2 2
1 1
1 1
4 5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20

Sample Output
4
465

题目链接：ZOJ-3955

题目大意：给出一个n*m的矩阵，定义矩阵中的特殊位置(i, j)为当且仅当Aij 是这一行唯一最小的数，是这一列唯一最大的数。删去一些行和列，剩下的元素构成的矩阵一共有(2^n-1)*(2^m-1)种，求这些矩阵的特殊位置的数量之和。

题目思路：我们可以算每个数对答案的贡献。设第 i 行比a[i][j]大的数有 k1 ，设第 j 列比a[i][j]小的数有 k2 个。则a[i][j]对答案的贡献就是 2^k1*2^k2

参考博客：here

以下是代码：

#include <iostream>#include <iomanip>#include <fstream>#include <sstream>#include <cmath>#include <cstdio>#include <cstring>#include <cctype>#include <algorithm>#include <functional>#include <numeric>#include <string>#include <set>#include <map>#include <stack>#include <vector>#include <queue>#include <deque>#include <list>using namespace std;#define LL long long#define mod 1000000007LL qpow(LL a, LL n){    LL ret = 1;    while (n)    {        if (n & 1) ret = ret * a % mod;        a = a * a % mod;        n>>= 1;    }    return ret;}long long num[1005][1005];long long row[1005][1005];long long col[1005][1005];int main() {    int t;    cin >> t;    while(t--)    {        int n,m;        scanf("%d%d",&n,&m);        for (int i = 0; i < n; i++)        {            for (int j = 0; j < m; j++)            {                scanf("%lld",&num[i][j]);                row[i][j] = num[i][j];                col[j][i] = row[i][j];            }        }        for (int i = 0; i < n; i++) sort(row[i], row[i] + m);        for (int i = 0; i < m; i++) sort(col[i], col[i] + n);        long long ans = 0;        for (int i = 0; i < n; i++)        {            for (int j = 0; j < m; j++)            {                long long a = upper_bound(row[i], row[i] + m, num[i][j]) - row[i];                long long b = lower_bound(col[j], col[j] + n, num[i][j]) - col[j];                a = m - a;                ans = (ans + qpow(2,a) * qpow(2,b) % mod) % mod;            }        }        printf("%lld\n",ans);    }    return 0;}