离散事件模拟-银行管理——队列思想（双队列）

你的名字 2022-06-12 03:08 217阅读 0赞

Think:  
1知识点：队列+模拟  
2反思：  
1、心态反思：实训结束不久，心态松懈，基础知识方面多次犯错，分析问题不深入、不全面  
2、知识点反思：基础知识不牢固，关于运算符优先级方面知识需要提高  
[运算符优先级——百度百科][Link 1]  
3、错误反思：不同时刻窗口的人数不一定相等  
3题目分析：双队列模拟银行的双窗口，注意平均逗留时间的计算

以下为Wrong Answer代码——不同时刻窗口的人数不一定相等

#include <bits/stdc++.h>
    
    using namespace std;
    
    struct node {
        int Begin;
        int End;
    }link_1, link_2;
    int tp1, tp2;
    
    int main(){
        int T, n, i, x, y;
        scanf("%d", &T);
        while(T--){
            scanf("%d", &n);
    
            double sum = 0;
            tp1 = tp2 = 0;
            link_1.Begin = link_1.End = 0;
            link_2.Begin = link_2.End = 0;
    
            for(i = 0; i < n; i++){
                scanf("%d %d", &x, &y);
                if(x < 0 || x > 360){
                    continue;
                }
                if(tp1 <= tp2){
                    if(tp1 == 0){
                        link_1.Begin = x;
                        link_1.End = x + y;
                    }
                    else {
                        if(x >= link_1.End){
                            sum += (double)(link_1.End - link_1.Begin);
                            link_1.Begin = x;
                            link_1.End = x + y;
                        }
                        else {
                            sum += (double)(link_1.End - link_1.Begin);
                            sum += (double)(link_1.End - x);
                            link_1.Begin = link_1.End;
                            link_1.End += y;
                        }
                        tp1--;
                    }
                    tp1++;
                }
                else {
                    if(tp2 == 0){
                        link_2.Begin = x;
                        link_2.End = x + y;
                    }
                    else {
                        if(x >= link_2.End){
                            sum += (double)(link_2.End - link_2.Begin);
                            link_2.Begin = x;
                            link_2.End = x + y;
                        }
                        else {
                            sum += (double)(link_2.End - link_2.Begin);
                            sum += (double)(link_2.End - x);
                            link_2.Begin = link_2.End;
                            link_2.End += y;
                        }
                        tp2--;
                    }
                    tp2++;
                }
            }
    
            sum += (double)(link_1.End - link_1.Begin);
            sum += (double)(link_2.End - link_2.Begin);
    
            printf("%.2lf\n", sum/(n));
        }
        return 0;
    }
    
    
    /***************************************************
    User name: 
    Result: Wrong Answer
    Take time: 0ms
    Take Memory: 224KB
    Submit time: 2017-07-13 20:34:04
    ****************************************************/

以下为Accepted代码

#include <bits/stdc++.h>
    
    using namespace std;
    
    struct node {
        int Begin;
        int End;
    }link_1[104], link_2[104];
    int op1, op2, tp1, tp2;
    
    int main(){
        int T, n, i, x, y;
        scanf("%d", &T);
        while(T--){
            op1 = op2 = tp1 = tp2 = 0;
            double sum = 0;
            memset(link_1, 0, sizeof(link_1));
            memset(link_2, 0, sizeof(link_2));
    
            scanf("%d", &n);
            for(i = 0; i < n; i++){
                scanf("%d %d", &x, &y);
                while(op1 < tp1 && x >= link_1[op1].End){
                    op1++;
                }
                while(op2 < tp2 && x >= link_2[op2].End){
                    op2++;
                }
                if(op1 == tp1){
                    link_1[tp1].Begin = x;
                    link_1[tp1].End = x + y;
                    tp1++;
                    sum += (double)(y);
                }
                else if(op2 == tp2){
                    link_2[tp2].Begin = x;
                    link_2[tp2].End = x + y;
                    tp2++;
                    sum += (double)(y);
                }
                else if((tp1 - op1) <= (tp2 - op2)){
                    link_1[tp1].Begin = link_1[tp1-1].End;
                    link_1[tp1].End = link_1[tp1-1].End + y;
                    sum += (double)(link_1[tp1].End - x);
                    tp1++;
                }
                else if((tp1 - op1) > (tp2 - op2)){
                    link_2[tp2].Begin = link_2[tp2-1].End;
                    link_2[tp2].End = link_2[tp2-1].End + y;
                    sum += (double)(link_2[tp2].End - x);
                    tp2++;
                }
            }
            printf("%.2lf\n", sum/n);
        }
        return 0;
    }
    
    
    /***************************************************
    User name: 
    Result: Accepted
    Take time: 0ms
    Take Memory: 228KB
    Submit time: 2017-07-13 21:15:51
    ****************************************************/

[Link 1]: http://baike.baidu.com/link?url=x9cgatZsEr62orMeWL_oS-5nBgrK4LOSRQ7PakPynnZ-MjhOhdT2U60YtsnMJOuKpFswaenWAxGiIvsnfUW6nDqdI6hHx1m9-HEfUYxfp1swcKsN1JE0UvscG4oCXZKhW9ERQ8ssgNRGbMOMpxpeja