3218力扣 ：切蛋糕的最小总开销1

class Solution {
public:
    int minimumCost(int m, int n, vector<int>& horizontalCut, vector<int>& verticalCut) {
        int x=1,y=1;//x代表纵向看蛋糕的数量 y代表横向看蛋糕的数量
        int res=0;
        sort(horizontalCut.begin(),horizontalCut.end());
        sort(verticalCut.begin(),verticalCut.end());
        while(horizontalCut.size()>0 && verticalCut.size()>0){
            if(verticalCut.back()>=horizontalCut.back()){
                res+=y*verticalCut.back();
                verticalCut.pop_back();
                x++;
            }
            else{
                res+=x*horizontalCut.back();
                horizontalCut.pop_back();
                y++;
            }
        }
 
        while(verticalCut.size()>0){
                res+=y*verticalCut.back();
                verticalCut.pop_back();
                x++;
            }
        while(horizontalCut.size()>0){
                res+=x*horizontalCut.back();
                horizontalCut.pop_back();
                y++;
            }
        return res;
    }
};

思维题 先把单刀开销最大的切了