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CodeForces 711A Bus to Udayland (水题)
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;}
CodeForces 711A Bus to Udayland (水题)
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