CCF_炉石传说

该编程题主要考察的是数据结构的使用，一个好的数据结构处理问题来得心应手
我是通过利用C++中原有的list数据结构（大体上相当于一个双向链表）解决该问题的，其中list的每个元素需保存角色和随从的position、attack、health。

#include<iostream>#include<string>#include<list>using namespace std;class AI{public:    int position;    int attack;    int health;    AI(int p,int a,int h){        position=p;        attack=a;        health=h;    }};typedef list<AI> LISTAI;int main(){    int n;    cin>>n;    LISTAI pet_one,pet_two;    pet_one.push_front(AI(0,0,30));    pet_two.push_front(AI(0,0,30));    LISTAI::iterator ite;    string action;    LISTAI *pet_att=&pet_one;    LISTAI *pet_def=&pet_two;    for(int i=0;i<n;i++){        cin>>action;        if(action=="summon"){            int p,a,h,tag=0;            cin>>p>>a>>h;            AI temp(p,a,h);            for(ite=pet_att->begin();ite!=pet_att->end();++ite){                if(ite->position==p){                    pet_att->insert(ite,temp);                    tag=1;                    while(ite!=pet_att->end()){                        ite->position++;                        ite++;                    }                break;                }            }            if(tag!=1) pet_att->push_back(temp);        }        else if(action=="attack"){            int att,def;            cin>>att>>def;            LISTAI::iterator ite_att,ite_def;            for(ite_att=pet_att->begin();ite_att!=pet_att->end();++ite_att)                if(ite_att->position==att) break;            for(ite_def=pet_def->begin();ite_def!=pet_def->end();++ite_def)                if(ite_def->position==def) break;            ite_att->health-=ite_def->attack;            ite_def->health-=ite_att->attack;            if(ite_att->health<=0){                ite_att=pet_att->erase(ite_att);                for(;ite_att!=pet_att->end();++ite_att)                    ite_att->position--;            }            if(ite_def->position!=0&&ite_def->health<=0){                ite_def=pet_def->erase(ite_def);                for(;ite_def!=pet_def->end();++ite_def)                    ite_def->position--;            }            else if(ite_def->position==0&&ite_def->health<=0){                break;            }        }        else if(action=="end"){            LISTAI *pp=pet_att;            pet_att=pet_def;            pet_def=pp;        }    }    if(pet_one.begin()->health<=0) cout<<-1<<endl;    else if(pet_two.begin()->health<=0) cout<<1<<endl;    else cout<<0<<endl;    ite=pet_one.begin();    cout<<ite->health<<endl;    ite++;    cout<<pet_one.size()-1;    for(;ite!=pet_one.end();++ite){        cout<<" "<<ite->health;    }    cout<<endl;    ite=pet_two.begin();    cout<<ite->health<<endl;    ite++;    cout<<pet_two.size()-1;    for(;ite!=pet_two.end();++ite){        cout<<" "<<ite->health;    }    cout<<endl;    return 0;}