CodeForces 711A Bus to Udayland (水题)

时间：2023-11-25 20:18:20

题意：给定一个n*4的矩阵，然后O表示空座位，X表示已经有人了，问你是不能找到一对相邻的座位，都是空的，并且前两个是一对，后两个是一对。

析：直接暴力找就行。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <list>

#include <sstream>

#define frer freopen("in.txt", "r", stdin)

#define frew freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e3 + 5;

const int mod = 1e9 + 7;

const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};

const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

char s[maxn][10];

int main(){

    while(cin >> n){

        bool ok = false;

        for(int i = 0; i < n; ++i){

            scanf("%s", &s[i]);

            if(!ok && s[i][0] == 'O' && s[i][1] == 'O'){

                s[i][0] = s[i][1] = '+';

                ok = true;

            }

            else if(!ok && s[i][3] == 'O' && s[i][4] == 'O'){

                s[i][3] = s[i][4] = '+';

                ok = true;

            }

        }

        printf("%s\n", ok ? "YES" : "NO");

        if(ok)

            for(int i = 0; i < n; ++i)

                puts(s[i]);

    }

    return 0;

}

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