atcoder ABC 355-C题详解

news2025/2/28 17:17:46

atcoder ABC 355-C题详解

Problem Statement

There is an N×N grid, where the cell at the i-th row from the top and the j-th column from the left contains the integer N×(i−1)+j.

Over T turns, integers will be announced. On Turn i, the integer Ai is announced, and the cell containing Ai is marked. Determine the turn on which Bingo is achieved for the first time. If Bingo is not achieved within T turns, print -1.

Here, achieving Bingo means satisfying at least one of the following conditions:

There exists a row in which all N cells are marked.
There exists a column in which all N cells are marked.
There exists a diagonal line (from top-left to bottom-right or from top-right to bottom-left) in which all N cells are marked.

Constraints

2≤N≤2×103
1≤T≤min(N2,2×105)
1≤Ai≤N2
Ai=Aj if i=j.
All input values are integers.

Input

The input is given from Standard Input in the following format:

Output

If Bingo is achieved within T turns, print the turn number on which Bingo is achieved for the first time; otherwise, print -1.

Sample Input 1

3 5
5 1 8 9 7

Sample Output 1

The state of the grid changes as follows. Bingo is achieved for the first time on Turn 4.

在这里插入图片描述

Sample Input 2

3 5
4 2 9 7 5

Sample Output 2

-1
Bingo is not achieved within five turns, so print -1.

Sample Input 3

4 12
13 9 6 5 2 7 16 14 8 3 10 11

Sample Output 3

思路分析：

先判断输入的数字在哪一行那一列，然后判断是否符合宾果游戏规则，判断主对角线和次对角线以及行、列。

code：

#include <iostream>
using namespace std;
const int N = 2010;
int n, t;
int row[N];
int col[N];
int ans1, ans2;
int main() {
    cin >> n >> t;
    for (int i = 0; i < t; i++) {
        int a;
        cin >> a;
        a--;//要减一比如5，5/3=1,5%3=2
        int row1 = a / n;
        int col1 = a % n;
//        if (row1 >= n || col1 >= n) {不需要比如第一个样例9/3=3 直接输出-1结束程序了
//            cout << "-1" << endl;
//            return 0;
//        }
        row[row1]++;
        col[col1]++;
        // 检查主对角线
        if (row1 == col1) {
         ans1++;
        }
        // 检查次对角线
        if (row1 + col1 == n - 1) {
         ans2++;
        }
        if(ans1 == n || ans2 == n) {
         cout << i + 1 << endl;
         return 0;
  }
        // 检查行和列
        if (row[row1] == n || col[col1] == n) {
            cout << i + 1 << endl;
            return 0;
        }
    }
    cout << "-1" << endl;
    return 0;
}