Given an m * n matrix M initialized with all 0's and several update operations.

Operations are represented by a 2D array, and each operation is represented by an array with two positive integers a and b, which means M[i][j] should be added by one for all 0 <= i < a and 0 <= j < b.

You need to count and return the number of maximum integers in the matrix after performing all the operations.

Example 1:Input: m = 3, n = 3operations = [[2,2],[3,3]]Output: 4Explanation: Initially, M = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]After performing [2,2], M = [[1, 1, 0], [1, 1, 0], [0, 0, 0]]After performing [3,3], M = [[2, 2, 1], [2, 2, 1], [1, 1, 1]]So the maximum integer in M is 2, and there are four of it in M. So return 4.Note:The range of m and n is [1,40000].The range of a is [1,m], and the range of b is [1,n].The range of operations size won't exceed 10,000.

题目大意：给出矩阵大小，给出每次加的坐标(x,y)那么每次操作都会使(0~x,0~y)这个矩阵内的格子加1，求最大的格子的个数。

class Solution {public:    int maxCount(int m, int n, vector
    
     
      >& ops) {        vector
      
       
         > v;        if (ops.size() == 0) return m*n;        int x = 100000;        int y = 100000;        for (int i = 0; i < ops.size(); ++i) {            x = min(x, ops[i][0]);            y = min(y, ops[i][1]);        }        return x*y;    }};