Problem L Tic-Tac-Toe
Accept: 94 Submit: 184
Time Limit: 1000 mSec Memory Limit : 262144 KB
Problem Description
Kim likes to play Tic-Tac-Toe.
Given a current state, and now Kim is going to take his next move. Please tell Kim if he can win the game in next 2 moves if both player are clever enough.
Here “next 2 moves” means Kim’s 2 move. (Kim move,opponent move, Kim move, stop).
Game rules:
Tic-tac-toe (also known as noughts and crosses or Xs and Os) is a paper-and-pencil game for two players, X and O, who take turns marking the spaces in a 3×3 grid. The player who succeeds in placing three of their marks in a horizontal, vertical, or diagonal row wins the game.
Input
First line contains an integer T (1 ≤ T ≤ 10), represents there are T test cases.
For each test case: Each test case contains three lines, each line three string(“o” or “x” or “.”)(All lower case letters.)
x means here is a x
o means here is a o
. means here is a blank place.
Next line a string (“o” or “x”) means Kim is (“o” or “x”) and he is going to take his next move.
Output
For each test case:
If Kim can win in 2 steps, output “Kim win!”
Otherwise output “Cannot win!”
Sample Input
3. . .. . .. . .oo x oo . xx x oxo x .. o .. . xo
Sample Output
Cannot win!Kim win!Kim win!
九宫棋Kim先下两步之内是否可以胜利
题解下次贴
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <cmath>
using namespace std;
char map[][];
int s[][],t,sum,k,cnt;
bool judge(int x,int y){
if((x+y)%==){//当此时这个格子的行列的和为奇数时
cnt=,sum=; //cnt代表可能胜利的次数
for(int i=;i<=;i++){
sum+=s[i][y];
}
sum*=k; //同是负数相乘为正数
if(sum==) //空白数为2时代表可以
return true;
else if(sum==) //当sum和为1时,此时可能胜利
cnt++;
sum=; //注意,此时一定要重置sum,因为sum在这个函数的前面后面的含义不同
for(int i=;i<=;i++){
sum+=s[x][i];//计算此时画相同的符号的个数之和
}
sum*=k;
if(sum==) //如果画相同符号的和为2,代表画下一个一定会胜利
return true;
else if(sum==) //如果此时画的相同的符合为1,代表可能胜利
cnt++;
if(cnt==){ //当可能胜利的次数超过2时一定可以胜利
return true;
}
}
else { //当此时这个各自的行列的和为偶数时
cnt=,sum=;
for(int i=;i<=;i++){
sum+=s[i][y];
}
sum*=k;
if(sum==)
return true;
else if(sum==)
cnt++;
sum=;
for(int i=;i<=;i++){
sum+=s[x][i];
}
sum*=k;
if(sum==)
return true;
else if(sum==)
cnt++;
sum=;
if(x==y){ //代表这个空白所在的地方在斜线上
for(int i=;i<=;i++){
sum+=s[i][i];
}
sum*=k; if(sum==)
return true;
else if(sum==)
cnt++;
}
else {
for(int i=;i<=;i++){
sum+=s[i][-i];
}
sum*=k; if(sum==)
return true;
else if(sum==)
cnt++;
}
if(cnt>=){
return true;
}
}
return false;
} int main(){
char st;
int flag;
scanf("%d",&t);
while(t--){
flag=;
memset(s,,sizeof());
for(int i=;i<=;i++){
for(int j=;j<=;j++){
cin>>map[i][j];
if(map[i][j]=='.') s[i][j]=;
else if(map[i][j]=='o') s[i][j]=;
else if(map[i][j]=='x') s[i][j]=-;
}
}
cin>>st; //代表此时Kim所用的符号
if(st=='o') k=;
else if (st=='x') k=-;
for(int i=;i<=;i++){
for(int j=;j<=;j++){
if(map[i][j]=='.'){
if(judge(i,j))
flag=;
}
}
}
if(flag) puts("Kim win!");
else puts("Cannot win!");
}
return ;
}