poj 1013

不看网上的discuss我还真想不出这个思路：
1) 平衡状态，所有参与称重的硬币都是真的；
2) 非平衡状态，所有没有参与称重的硬币都是真的；
3) 若有一枚硬币既出现在重的一边，又出现在轻的一边，则这枚硬币是真的；
经过以上3重筛选，剩下的硬币恰好1枚，否则答案不唯一。

code：#include <iostream>#include <string>#include <vector>#include <map>using namespace std;int main(){string letters = "ABCDEFGHIJKL";vector<char> fake;vector<string> weight;string light, heavy;string s1, s2, state;int cases;cin>>cases;while(cases--){heavy.clear();light.clear();map<char, bool> result;for(int i = 0; i < 12; ++i){result.insert(make_pair('A'+i,0));}for(int ix = 0; ix < 3; ++ix){cin>>s1>>s2>>state;if(state == "even"){for(size_t ix = 0; ix != s1.size(); ++ix){result[s1[ix]] = 1;result[s2[ix]] = 1;}}if(state == "up" || state == "down"){string s3 = s1 + s2;string::size_type pos1 = 0;while((pos1 = letters.find_first_not_of(s3,pos1)) != string::npos){result[letters[pos1]] = 1;++pos1;}if(state == "up"){heavy.append(s1);light.append(s2);}if(state == "down"){heavy.append(s2);light.append(s1);}}}string::size_type pos = 0;while((pos = heavy.find_first_of(light,pos)) != string::npos){result[heavy[pos]] =1;++pos;}for(map<char, bool>::iterator iter = result.begin(); iter != result.end(); ++iter){if(iter->second == 0){fake.push_back(iter->first);if(heavy.find(iter->first) != string::npos){weight.push_back("heavy");}else if(light.find(iter->first) != string::npos){weight.push_back("light");}break;}}}for(size_t in = 0; in != fake.size(); ++in){cout<<fake[in]<<" is the counterfeit coin and it is "<<weight[in]<<"./n";}return 0;}

main gains：

1 string的库函数：find_first_of, find_first_not_of, find等，很好用，注意当不存在时返回string::npos，不要用size()，会导致边界溢出！

2 变量在循环中重复应用时应严重注意需不需要清空！