力扣 第 208 场周赛

今晚剩一个小时左右了emmmm能写多少写多少哈,

并且小白做题慢，还要参考大佬的题解优化emmmmm肯定做不完呐呐呐（我才不会说是因为我不会做呢)

那，就开始咯～

 模拟一下栈的思路，，还行吧

就是，，一开始想多了，想匹配所有字符emmmm关键/还涉及转制

俺还写了测试主函数emmmm

#include <iostream>
#include <stdio.h>
#include <stack>
#include <vector>
#include <math.h>
#include <string>
#include <algorithm>
#include <unordered_map>
#include <map>
#include<set>
#include <cstring>
#include <queue>

#include <stdio.h>
#include <stdlib.h>
#define IOS ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)

#define ll long long

using namespace std;

class Solution {
public:
    int minOperations(vector<string>& logs) {
        stack<int> st;
        for (int i = 0; i < logs.size(); ++i) {
            if(logs[i][0]=='.' && logs[i][1]=='.'){
                while(!st.empty())
                    st.pop();
            }
            else if(logs[i][0]=='.'){

            }
            else{
                st.push(1);
            }

        }
        return st.size();
    }
};
int main(){
    Solution s;
    vector<string> v={"d1/","d2/","../","d21/","./"
    };
    cout<<s.minOperations(v);
}

　　

 读题读了半天，幸亏不是英文的

就是个模拟嘛，然而心态在这一刻还是崩溃了

具体是为啥呢，emmmmm看了看样例发现，其实在中间有可能游客数为0后面还有人

我限制写少了...

 这就对了呐呐

我写的main函数检测第130个样例，，，，，，

大佬有手就行的题本小白得研究好久........

#include <iostream>
#include <stdio.h>
#include <stack>
#include <vector>
#include <math.h>
#include <string>
#include <algorithm>
#include <unordered_map>
#include <map>
#include<set>
#include <cstring>
#include <queue>

#include <stdio.h>
#include <stdlib.h>
#define IOS ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)

#define ll long long

using namespace std;
class Solution {
public:
    int minOperationsMaxProfit(vector<int>& customers, int boardingCost, int runningCost) {
        int res = 0;
        int finres = 0;
        int times = 0;
        int fintimes = 0;
        int num = 0;
        int s=customers.size();
        for (int i=0;;i++) {
            if(i!=0 && num == 0)
                break;
            times++;
            if(i<s) {
                num += customers[i];
            }
            if (num >= 4) {
                res += (4 * boardingCost - runningCost);
                num -= 4;
            }
            else {

                res += (num * boardingCost - runningCost);
                num = 0;
            }

            if (res > finres) {
                finres = res;
                fintimes = times;
            }
            cout<<finres<<" "<<fintimes<<endl;
        }

        return finres ? fintimes : -1;

    }

};
int main(){
    Solution s;
    vector<int> v={2,16,22,15,46,10,23,0,13,23,33,14,21,19,47,6,13,28,37,18,7,44,25,11,35,10,12,37,27,16,18,48,41,23,22,14,50,39,14,15,26,9,6,13,24,18,36,50,16,33,13,27,11,37,25,18,21,39,41,1,47,44,39,8,40,40,8,0,5,25,44,24,0,27,15,1
    };
    int a=88,b=6;
    cout<<s.minOperationsMaxProfit(v,a,b);
}

困了，不知能撑多久

第三题不想看了，太长了，等我咕咕咕

 这网速有点不太好

class Solution {
    int max_satisfied = 0;
public:
    int maximumRequests(int n, vector<vector<int>>& requests) {
        vector<int> count(n, 0);
        dfs(requests, count, 0, 0);
        return max_satisfied;
    }
    void dfs(vector<vector<int>>& requests, vector<int>& count, int idx, int satisfied)
    {
        if(satisfied+requests.size()-idx <= max_satisfied)
            return;//剩余的假如都可以满足，都不可能超过最大的，剪枝
        if(idx == requests.size())
        {
            for(int i = 0; i < count.size(); i++)
            {
                if(count[i] != 0)
                    return;
            }
            max_satisfied = max(max_satisfied, satisfied);
            return;
        }

        int from = requests[idx][0], to = requests[idx][1];
        count[from]--;
        count[to]++;
        dfs(requests, count, idx+1, satisfied+1);//选
        count[from]++;
        count[to]--;
        dfs(requests, count, idx+1, satisfied);//当前不选
    }
};

呐呐呐网速不想让我写了，待我回来更三．