【UVA】230 - Borrowers（map模拟）

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利用map<string,int>判断一本书的状态，0代表借出去了，1代表在书架，2代表借出去换回来但是还未放回书架

设计一些字符串的处理问题，用一些字符串搜索函数比如 strstr , strchar等等

14072706230BorrowersAcceptedC++0.0152014-08-21 02:59:27

AC代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<stack>#include<queue>#include<map>#include<set>#include<list>#include<cmath>#include<string>#include<sstream>#include<ctime>using namespace std;#define _PI acos(-1.0)#define INF (1 << 10)#define esp 1e-9typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> pill;/*======================================================================================*/#define MAXD 1000 + 10#define LEN 80 + 10struct Book{    char name[LEN];    char author[LEN];    friend bool operator < (Book p , Book q){        int e1 = strcmp(p.author,q.author);        int e2 = strcmp(p.name,q.name);        if(e1 != 0){            if(e1 > 0)                return false;            else                return true;        }        else{            if(e2 > 0)                return false;            else                return true;        }    }}book[MAXD];int n = 0;map<string,int>value;int find_pre(int cur){    for(int i = cur - 1 ; i >= 0 ; i--){         if(value[book[i].name] == 1)            return i;    }    return -1;}int main(){    char str[LEN];    int  pos;    value.clear();    while(gets(str)){        if(strcmp(str,"END") == 0)            break;        pos = strchr(str + 1, '"') - str;        strncpy(book[n].name,str,pos + 1);        book[n].name[pos + 2] = '\0';        pos = strstr(str + pos , "by") - str;        strcpy(book[n].author,str + pos + 3);        value[book[n].name] = 1;        n++;    }    sort(book , book + n);    while(scanf("%s",str)){        if(strcmp(str,"END") == 0)            break;        if(strcmp(str,"BORROW") == 0){             gets(str);             value[str + 1] = 0;        }        else if(strcmp(str,"RETURN") == 0){             gets(str);             value[str + 1] = 2;        }        else if(strcmp(str,"SHELVE") == 0){             for(int i = 0 ; i < n ; i++)                if(value[book[i].name] == 2){                      pos = find_pre(i);                      if(pos == -1)                        printf("Put %s first\n",book[i].name);                      else                        printf("Put %s after %s\n",book[i].name,book[pos].name);                      value[book[i].name] = 1;                }             printf("END\n");        }    }    return 0;}

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