题解 | 贪吃蛇游戏

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

deque<pair<int,int>> snake;
set<pair<int,int>>s;

bool moveSnake(int dir) {
    // TODO: 请实现移动逻辑
         set<pair<int,int>>s;//检查碰撞的临时容器
         pair<int,int>new_head=snake.back();
        if(dir==1){
       new_head.second++;
        }
        if(dir==2){
        new_head.second--;
        }
        if(dir==3){
           new_head.first--;
        }
        if(dir==4){
          new_head.first++;    
        }//计算蛇头
        snake.push_back(new_head);//加新头
         snake.pop_front();//删尾巴  
        for(int i=0;i<snake.size();i++){
            s.insert(snake[i]);
        }//碰撞检测 存在重复set会更小
        if(s.size()!=snake.size()){
            return true;
        }
       
        return false;

}

bool eatSnake(int dir) {
    // TODO: 请实现吃果子生长逻辑
     set<pair<int,int>>s;//检查碰撞的临时容器
         pair<int,int>new_head=snake.back();
        if(dir==1){
       new_head.second++;
        }
        if(dir==2){
        new_head.second--;
        }
        if(dir==3){
           new_head.first--;
        }
        if(dir==4){
          new_head.first++;    
        }//计算蛇头
        snake.push_back(new_head);//加新头
        // snake.pop_front();//删尾巴 
        for(int i=0;i<snake.size();i++){
            s.insert(snake[i]);
        }//碰撞检测 存在重复set会更小
        if(s.size()!=snake.size()){
            return true;
        }
       
        return false;
}



int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, q;
    cin >> n >> q;
    snake.clear();
    for (int i = 0; i < n; i++) {
        int x, y;
        cin >> x >> y;
        snake.emplace_back(x, y);
    }
    for (int i = 0; i < q; i++) {
        int op, dir;
        cin >> op >> dir;
        bool collision = (op == 1 ? moveSnake(dir) : eatSnake(dir));
        if (collision) {
            cout << -1 << '\n';
            return 0;
        } else {
            for (auto it = snake.rbegin(); it != snake.rend(); ++it) {
                cout << it->first << ' ' << it->second << '\n';
            }
        }
    }
    return 0;
}