首页 > 代码库 > 【Lytning Test1】 资源管理器
【Lytning Test1】 资源管理器
这题是道大模拟= =
代码比较长，并且需要注意的细节很多，各种不合法的命令判定简直巨坑= =
还有一些容易忽视的地方， 比如 new 操作， 竟然会有 new 1 或 new 0 这种丧心病狂的命令， 结果我刚开始的判定方法就崩了= =
最后不得不改成读入一行命令再切割。
我觉得最麻烦的操作就是 back 和 fore 了，刚开始想不明白是什么样子的，建议先做一下 POJ 1028 ，一道模拟浏览器的题。
代码如下：
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <cstdlib>#include <vector>#include <stack>using namespace std;int n, historyTop = 0, historyNum = 0;char Cmd[50 + 5], cmd2[50 + 5], cmd3[50 + 5], Prefix[4048 + 10];struct File{ int Type, Size, Depth; char Name[50 + 5]; File *Fa; vector<File *> subFile;} F[2014 + 5], *P = F, *root, *now;stack<File *> foreStack, backStack;void solveNew(){ int fileType, spaceLoc; getchar(); gets(cmd2); spaceLoc = 0; while (spaceLoc <= strlen(cmd2) - 1 && cmd2[spaceLoc] != ‘ ‘) spaceLoc++; if (spaceLoc < strlen(cmd2) - 1) { if (cmd2[0] == ‘0‘ && spaceLoc == 1) fileType = 0; else if (cmd2[0] == ‘1‘ && spaceLoc == 1) fileType = 1; else { printf("WRONG COMMAND.\n"); return; } for (int i = 2; i <= strlen(cmd2) - 1; i++) cmd3[i - 2] = cmd2[i]; cmd3[strlen(cmd2) - 1 - 2 + 1] = ‘\0‘; } else { fileType = 0; strcpy(cmd3, cmd2); } for (int i = 0; i <= now -> Size - 1; i++) { if (!strcmp(cmd3, now -> subFile[i] -> Name) && now -> subFile[i] -> Type == fileType) { printf("WRONG COMMAND.\n"); return; } } File *newFile = ++P; newFile -> Type = fileType; newFile -> Size = 0; newFile -> Depth = now -> Depth + 1; strcpy(newFile -> Name, cmd3); newFile -> Fa = now; now -> Size++; now -> subFile.push_back(newFile); if (fileType == 1) { now = newFile; backStack.push(now); while (!foreStack.empty()) foreStack.pop(); } }void solveUp(){ if (now -> Fa == NULL) { printf("WRONG COMMAND.\n"); return; } now = now -> Fa; backStack.push(now); while (!foreStack.empty()) foreStack.pop();}void solveEnter(){ char fileName[50 + 5], tempName[50 + 5]; File *p = now; scanf("%s", cmd2); int loc, num; bool flag = false; while (strlen(cmd2)) { if (p -> Size == 0) { printf("WRONG COMMAND.\n"); return; } loc = 0; while (loc <= strlen(cmd2) - 1 && cmd2[loc] != ‘\\‘) loc++; if (loc == 0 || loc == strlen(cmd2)) { printf("WRONG COMMAND.\n"); return; } strncpy(fileName, cmd2, loc); fileName[loc] = ‘\0‘; flag = false; for (int i = 0; i <= p -> Size - 1; i++) { if (!strcmp(fileName, p -> subFile[i] -> Name) && p -> subFile[i] -> Type == 1) { flag = true; p = p -> subFile[i]; break; } } if (!flag) { printf("WRONG COMMAND.\n"); return; } if (loc == strlen(cmd2) - 1) break; for (int i = loc + 1; i <= strlen(cmd2) - 1; i++) tempName[i - loc - 1] = cmd2[i]; num = strlen(cmd2) - loc - 1; strncpy(cmd2, tempName, num); cmd2[num] = ‘\0‘; } now = p; backStack.push(now); while (!foreStack.empty()) foreStack.pop();}void solveBack(){ if (backStack.size() == 1) { printf("WRONG COMMAND.\n"); return; } foreStack.push(backStack.top()); backStack.pop(); now = backStack.top();}void solveFore(){ if (foreStack.size() == 0) { printf("WRONG COMMAND.\n"); return; } now = foreStack.top(); foreStack.pop(); backStack.push(now);}void solvePrint(File *p){ Prefix[p -> Depth * 2] = ‘\0‘; printf("%s", Prefix); Prefix[p -> Depth * 2] = ‘.‘; if (p -> Type == 1) printf("\\"); printf("%s\n", p -> Name); for (int i = 0; i <= p -> Size - 1; i++) solvePrint(p -> subFile[i]);}int main(){ freopen("explorer.in", "r", stdin); freopen("explorer.out", "w", stdout); scanf("%d", &n); root = ++P; root -> Type = 1; root -> Size = 0; root -> Depth = 0; strcpy(root -> Name, "exp"); root -> Fa = NULL; now = root; backStack.push(now); for (int i = 1; i <= 4048 + 5; i++) Prefix[i] = ‘.‘; for (int i = 1; i <= n; i++) { scanf("%s", Cmd); if (!strcmp(Cmd, "new")) solveNew(); if (!strcmp(Cmd, "up")) solveUp(); if (!strcmp(Cmd, "enter")) solveEnter(); if (!strcmp(Cmd, "back")) solveBack(); if (!strcmp(Cmd, "fore")) solveFore(); if (!strcmp(Cmd, "print")) solvePrint(root); } fclose(stdin); fclose(stdout); return 0;}
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。