CodeForces 3C---Tic-tac-toe--思维题

C. Tic-tac-toe

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

Certainly, everyone is familiar with tic-tac-toe game. The rules are very simple indeed. Two players take turns marking the cells in a 3 × 3grid (one player always draws crosses, the other — noughts). The player who succeeds first in placing three of his marks in a horizontal, vertical or diagonal line wins, and the game is finished. The player who draws crosses goes first. If the grid is filled, but neither Xs, nor 0s form the required line, a draw is announced.

You are given a 3 × 3 grid, each grid cell is empty, or occupied by a cross or a nought. You have to find the player (first or second), whose turn is next, or print one of the verdicts below:

• illegal — if the given board layout can't appear during a valid game;
• the first player won — if in the given board layout the first player has just won;
• the second player won — if in the given board layout the second player has just won;
• draw — if the given board layout has just let to a draw.

Input

The input consists of three lines, each of the lines contains characters ".", "X" or "0" (a period, a capital letter X, or a digit zero).

Output

Print one of the six verdicts: first, second, illegal, the first player won, the second player won or draw.

Sample test(s)

input

X0X
.0.
.X.

output

second

此题需考虑全面。

首先考虑不合法的情况，有如下4种：

1、X的数量少于0的数量。

2、X的数量-0的数量>1。

3、X数量等于0的数量，而此时存在三个X相连（即先手获胜）。这是不可能的，因为先手下子之后，X的数量必然多于0的数量，不可能相等。

4、X的数量大于0的数量，而此时存在三个0相连（即后手获胜）。这是不可能的，因为后手下子之后X的数量和0的数量应当相等。

注：当X为五个且有两个三连重叠时为第一个人赢，因为可能最后一步放在交叉点，--，一开始以为是illegal；

#include <iostream>

using namespace std;
char a[4][4];
bool sh(char c)
{
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<3;j++)
        {
            if(a[i][j]!=c)
                break;
            if(j==2)
                return 1;
        }
    }
    return 0;
}
bool sl(char c)
{
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<3;j++)
        {
            if(a[j][i]!=c)
                break;
            if(j==2)
                return 1;
        }
    }
    return 0;
}
bool sxl(char c)
{
    for(int i=0;i<3;i++)
    {
        if(a[i][i]!=c)
            break;
        if(i==2)
            return 1;
    }
    return 0;
}
bool sxr(char c)
{
    for(int i=0;i<3;i++)
    {
        if(a[i][2-i]!=c)
            break;
        if(i==2)
            return 1;
    }
    return 0;
}
int main()
{
    while(cin>>a[0])
    {   int sum1=0,sum2=0;
        cin>>a[1]>>a[2];
        for(int i=0;i<3;i++)
        {
            for(int j=0;j<3;j++)
            {
                if(a[i][j]=='X')
                    sum1++;
                if(a[i][j]=='0')
                    sum2++;
            }
        }
        int flag=0;
        if(sum1<sum2||sum1>sum2+1)
            flag=1;
        else if((sh('X')==1||sl('X')==1||sxl('X')==1||sxr('X')==1)&&sum1==sum2)
                flag=1;
        else if((sh('0')==1||sl('0')==1||sxl('0')==1||sxr('0')==1)&&sum1>sum2)
            flag=1;
        //else if((sh('X')==1&&sl('X')==1)||(sh('X')==1&&sxl('X')==1)||(sh('X')==1&&sxr('X')==1)||(sl('X')==1&&sxl('X')==1)||(sl('X')==1&&sxr('X')==1)||(sxr('X')==1&&sxl('X')==1))
            // flag=1;
        else if(sh('X')==1||sl('X')==1||sxl('X')==1||sxr('X')==1)
            flag=2;
        else if(sh('0')==1||sl('0')==1||sxl('0')==1||sxr('0')==1)
            flag=3;
        else if(sum1==sum2)
            flag=4;
        else if(sum1>sum2&&(sum1+sum2)!=9)
            flag=5;
        else if(sum1>sum2&&(sum1+sum2)==9)
            flag=6;
            if(flag==1)
                cout<<"illegal"<<endl;
            if(flag==2)
                cout<<"the first player won"<<endl;
            if(flag==3)
                cout<<"the second player won"<<endl;
            if(flag==4)
                cout<<"first"<<endl;
            if(flag==5)
                cout<<"second"<<endl;
            if(flag==6)
                cout<<"draw"<<endl;
    }
    return 0;
}