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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#547784#7517. Flying Ship StorySICKOWA 0ms3580kbC++174.7kb2024-09-05 10:10:482024-09-05 10:10:48

Judging History

你现在查看的是最新测评结果

  • [2024-09-05 10:10:48]
  • 评测
  • 测评结果:WA
  • 用时:0ms
  • 内存:3580kb
  • [2024-09-05 10:10:48]
  • 提交

answer

#include<iostream>
#define int long long
using namespace std;
struct poi{
    int x{}, y{}, v{};
}p[20], np[20];
int pin1, pin2;
signed main(){
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int n, res = 0; cin>>n;
    while(n--){
        int opt, x, y, v;
        cin>>opt;
        if(opt == 1){
            pin2=0;
            cin>>x>>y>>v; x ^= res; y^= res; v^= res;
            int lt = 1;
            int nx, ny;
            if(p[lt].v < v){
                nx = x; ny = y;
                np[++pin2] = {x, y, v}; v = -1;
            }
            else{
                nx = p[lt].x; ny = p[lt].y;
                np[++pin2] = p[lt++];
            }
            int lsx[4] = {0, 0, 0, 0}, lsy[4]={0, 0, 0, 0};
            while(lt <= pin1 || v > 0){
                poi pos;
                if(p[lt].v < v) {
                    pos = {x, y, v}; v = -1;
                }
                else{
                    pos = p[lt++];
                }
                // 阶段1 十字
                if(nx && ny){
                    // 无用
                    if(nx == pos.x && ny == pos.y) continue;
                    // 有用
                    np[++pin2] = pos;
                    // 阶段2
                    if(nx == pos.x && ny != pos.y){
                        ny = 0; continue;
                    }
                    if(nx != pos.x && ny == pos.y){
                        nx = 0; continue;
                    }
                    // 阶段3
                    if(nx != pos.x && ny != pos.y){
                        nx = ny = 0;
                        lsx[0] = lsy[0] = 2;
                        lsx[1] = nx; lsx[2] = pos.x;
                        lsy[1] = pos.y; lsy[2] = ny;
                        continue;
                    }
                }
                // 阶段2 单边的情况
                else if(nx || ny){
                    if(nx){
                        // 无用
                        if(pos.x == nx) continue;
                        // 有用
                        np[++pin2] = pos;
                        // 进入一个点状态
                        lsx[0] = lsy[0] = 1;
                        lsx[1] = nx; lsy[1] = pos.y;
                        nx = 0;
                        continue;
                    }
                    // 无用
                    if(pos.y == ny) continue;
                    // 有用
                    np[++pin2] = pos;
                    // 进入一个点状态
                    lsx[0] = lsy[0] = 1;
                    lsx[1] = pos.x; lsy[1] = ny;
                    ny = 0;
                    continue;
                }
                // 阶段3 点的情况
                else{
                    if(lsx[0] == 1){
                        // 无用
                        if(pos.x == lsx[1] || pos.y == lsy[1]) continue;
                        // 有用 全部结束
                        np[++pin2] = pos;
                        break;
                    }
                    if(lsx[0] == 2){
                        // 无用
                        if((pos.x == lsx[1] && pos.y == lsy[2]) || (pos.x == lsx[2] || pos.y == lsx[1])) continue;
                        // 有用
                        np[++pin2] = pos;
                        // 一个点
                        if((pos.x == lsx[2] || pos.y == lsy[2])){
                            // 进入一个点状态
                            lsx[0] = lsy[0] = 1;
                            lsx[1] = lsx[2]; lsy[1] = lsy[2];
                            continue;
                        }
                        if((pos.x == lsx[1] || pos.y == lsy[1])){
                            // 进入一个点状态
                            lsx[0] = lsy[0] = 1;
                            continue;
                        }
                        // 结束
                        break;
                    }
                }
            }
            for(int i=1; i<=pin2; i++) p[i] = np[i], np[i] = {0, 0, 0};
            pin1 = pin2;
            if(pin1 < 4){
                if(v > p[4].v){
                    pin1++; p[pin1] = {x, y, v};
                }
                if(p[4].v){
                    pin1++;
                }
            }
            for(int i=1; i<=2; i++) p[i+pin1] = {0, 0, 0};
        }
        if(opt == 2){
            res = 0;
            cin>>x>>y;  x ^= res; y ^= res;
            for(int i=1; i<=pin1; i++){
                if(p[i].x == x || p[i].y == y)
                    continue;
                res = p[i].v;
                break;
            }
            cout<<res<<"\n";
        }
    }
    return 0;
}

详细

Test #1:

score: 0
Wrong Answer
time: 0ms
memory: 3580kb

input:

5
1 2 3 1
1 4 5 2
2 2 2
2 3 7
2 3 4

output:

2
2
2

result:

wrong answer 2nd lines differ - expected: '1', found: '2'