CCF炉石传说---谁说玩游戏对学习没帮助的？

CCF-201609-3

我们先来看一下题目

解题思路

大模拟题，他怎么说就怎么写
不理解题意的自己去搓两把炉石传说就懂了！
存储随从用vector，这个增删随从方便呀！

代码

#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
struct node{
    int attack;
    int blood;
};
vector<node> a,b;
int A=30,B=30;
int turn=1;
int main(){
    int n;
    cin>>n;
    while(n--){
      string order;
      cin>>order;
      node t;
      if(order[0]=='s'){
          int pos;
          cin>>pos>>t.attack>>t.blood;
          if(turn==1){
            if(pos>a.size()) a.push_back(t);
            else a.insert(a.begin()+pos-1,t); 
          }else{
            if(pos>b.size()) b.push_back(t);
            else b.insert(b.begin()+pos-1,t);     
          }
      }else if(order[0]=='a'){
          int attacker,defender; 
          cin>>attacker>>defender;
          if(turn==1){ //A 攻击 B 
            if(defender!=0){
                a[attacker-1].blood-=b[defender-1].attack;
                b[defender-1].blood-=a[attacker-1].attack;
                if(a[attacker-1].blood<=0) a.erase(a.begin()+attacker-1);
                if(b[defender-1].blood<=0) b.erase(b.begin()+defender-1);
            }else{
                B-=a[attacker-1].attack;
                if(B<=0) break;
            }
          }else{       //B 攻击 A 
            if(defender!=0){
                b[attacker-1].blood-=a[defender-1].attack;
                a[defender-1].blood-=b[attacker-1].attack;
                if(b[attacker-1].blood<=0) b.erase(b.begin()+attacker-1);
                if(a[defender-1].blood<=0) a.erase(a.begin()+defender-1);
            }else{
                A-=b[attacker-1].attack;
                if(A<=0) break;
            }
          } 
     }else if(order[0]=='e'){
       turn*=-1;
     }
 }
 if(A<=0) cout<<-1<<endl;
 else if(B<=0) cout<<1<<endl;
 else cout<<0<<endl;
 cout<<A<<endl;
 cout<<a.size()<<" ";
 for(int i=0;i<a.size();i++){
     cout<<a[i].blood<<" ";
 }
 cout<<endl<<B<<endl;
 cout<<b.size()<<" ";
 for(int i=0;i<b.size();i++){
     cout<<b[i].blood<<" "; 
 }
 cout<<endl;
 return 0;
}