UVA1596 Bug Hunt

来源：互联网发布：mac电脑怎么压缩文件编辑：程序博客网时间：2024/06/01 09:12

这个题竟然卡了我一整天，最开始的思路是把每个数组元素都替换成数字并依次判断，果断写乱了，改了大半天都没改对。

今天换了个思路，把每一行代码的括号都去掉，仅保留字母和数字，这样进行判断会方便一些。同时，把每一行的字母（即数组名称提取出来），用循环过一遍，跟相应的数字对应，判断越界等的情况。

#include<iostream>#include<string>#include<vector>#include<map>#include<cstdio>#include<cstring>#include<cctype>#include<sstream>using namespace std;typedef long long ll;pair<char,ll>ha;  //记录数组的下标及名称map<char,ll>Count; //记录每个数组名的最大值map<pair<char,ll>,ll>num; //记录每个数组单元的值string trans(string s){   //把中括号去掉   char ss[10000];   memset(ss,0,sizeof(ss));   int i, j = 0, len = s.length();   for(i = 0; i < s.length(); i++)      if(s[i] != '[' && s[i] != ']') ss[j++] = s[i];   ss[j] = 0;   string s1 = ss;   return s1;}ll get_count(string s){      //得到数组元素的值并判断   int len = s.length(), temp = 0;   string s1 = s;   for(int i = 0; i < len; i++){      char c = s[i];      if(isalpha(c)) if(!Count.count(c)) return -1;   //没有定义的数组，直接返回      if(isalpha(c)) s[i] = ' ';      if(isdigit(c)) {temp = i;break;}   }   stringstream ss(s);   ll x;   ss >> x;   for(int i = temp-1; i >= 0; i--){      char c = s1[i];      ha.first = c; ha.second = x;      if(!num.count(ha)) return -1;      x = num[ha];   }   return x;}int main(){    string s;    int i,j,k;    vector<string>code;    while(cin >> s && s[0] != '.'){        Count.clear();        num.clear();        code.clear();        code.push_back(trans(s));        int yes = 1;        while(cin >> s && s[0] != '.')            code.push_back(trans(s));        int len = code.size();        for(i = 0; i < len; i++){            k = code[i].find('=');            if(k == std::string::npos){               string s1 = code[i].substr(1);               char c = code[i][0];               ll x = get_count(s1);               if(x == -1) {cout << i+1 << endl;yes = 0;break;}               Count[c] = x;            }            else{               string s1 = code[i].substr(1,k-1);               string s2 = code[i].substr(k+1);               char c = code[i][0];               ll x = get_count(s1), y = get_count(s2);               if(!Count.count(c) || x == -1 || y == -1 || x >= Count[c]){cout << i+1 << endl; yes = 0; break;}               ha.first = c; ha.second = x;               num[ha] = y;            }        }if(yes == 1)cout << "0\n";    }    return 0;}

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