CSP201609-3炉石传说

 

 

 

简单思路：

　　我们把每一个角色都看作一个节点，有攻击有血量，然后每一个玩家都维护一个vector就可以了，然后根据操作类型（3)按照题目要求来写出对应过程即可，这道题思路还是简单的。

代码：

#include<iostream>
#include<vector>
#include<string>
#include<cstdlib>
 
using namespace std;

struct node {
    int hth;
    int atk;
    node(int h, int a) : hth(h), atk(a) {}
};

vector<node> player[2];

int main(){
    
    int N;
    cin >> N;
    int pl = 0;
    player[0].push_back(node(30, 0));
    player[1].push_back(node(30, 0));
    
    for(int n=0; n<N; n++) {
        string options;
        cin >> options;
        if(options == "summon") {
            int pos, h, a;
            cin >> pos >> a >> h;
            player[pl].insert(player[pl].begin()+pos, node(h, a));
        }else if(options == "atk") {
            int att, deff;
            cin >> att >> deff;
            player[pl][att].hth -= player[!pl][deff].atk;
            player[!pl][deff].hth -= player[pl][att].atk;
            if(player[pl][att].hth <= 0) {
                player[pl].erase(player[pl].begin()+att);
            }
            if(player[!pl][deff].hth <= 0 && deff!=0) {
                player[!pl].erase(player[!pl].begin()+deff);
            }
        }else if(options == "end") {
            pl = !pl;
        }
    }
    if(player[0][0].hth>0 && player[1][0].hth>0) cout << 0 << endl;
    else if(player[0][0].hth>0) cout << 1 << endl;
    else cout << -1 << endl;
    for(int i=0; i<2; i++) {
        cout << player[i][0].hth << endl;
        cout << player[i].size()-1 << " ";
        for(int j=1; j<player[i].size(); j++) {
            cout << player[i][j].hth << " ";
        }
        cout << endl;
    }
}