Bus to Udayland

                          Bus to Udayland
                   time limit per test  2 seconds
                   memory limit per test  256 megabytes

ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has nrows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.

ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit?

Input
The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the number of rows of seats in the bus.

Then, n lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals ‘|’) and the last two of them denote the second pair of seats in the row.

Each character, except the walkway, equals to ‘O’ or to ‘X’. ‘O’ denotes an empty seat, ‘X’ denotes an occupied seat. See the sample cases for more details.

Output
If it is possible for Chris and ZS to sit at neighbouring empty seats, print “YES” (without quotes) in the first line. In the next n lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters ‘+’. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to ‘O’ in the input and to ‘+’ in the output).

If there is no pair of seats for Chris and ZS, print “NO” (without quotes) in a single line.

If there are multiple solutions, you may print any of them.

给你n排座位，其中|表示过道，过道两边每边有两个位子，现在给你一个当前状态，判断是否有一个连坐（二连坐）让我们来选择。

其中O表示空位，X表示有人了。

输出任意一个可行解。

思路：直接模拟



#include<iostream>
using namespace std;
char s [1005][6];
int main()
{
	int n,i,j,flag=0;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=5;j++)
		{
			cin>>s[i][j];
		}
	}
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=5;j++)
		{	if(s[i][j]=='O'&&s[i][j+1]=='O')
			{
				printf("YES\n");
				s[i][j]='+';
				s[i][j+1]='+';
				flag=1;
				break;
			}
		}if(flag==1)break;
	}
	if(flag==1)
	{
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=5;j++)
		{
			cout<<s[i][j];
			if(j==5)
				cout<<endl;
		}
	}
	}
	else
		printf("NO\n");
	return 0;
}

猜你喜欢