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.
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.
For each test case:
If Kim can win in 2 steps, output “Kim win!”
Otherwise output “Cannot win!”
3 . . . . . . . . . o o x o o . x x x o x o x . . o . . . x oSample Output
Cannot win! Kim win! Kim win!
题意:给出一个3x3的棋盘,两人用'x'和'o'下棋,三个棋子连成一条线就算赢,现在给出Kim的棋子样式,问Kim能否在两步之内取得胜利(若没有一步取胜,对手也会再下一步)。
分析:我们可以遍历棋盘上的点,若点为'.'则将其变为Kim的棋子进行一次判断,是否存在三子连珠,若存在则说明Kim可以赢得比赛。如果改变点后未形成三子连珠,则再次进行遍历,将'.'变为Kim的棋子,判断是否有两个以上的点符合三子连珠,若存在则Kim可以赢得比赛。
因为一步之后对手会下一步棋,如果只存在一种满足的情况,那么对手会将路堵死,所以要有两点以上满足条件。
代码如下:
#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
const int MX = 5;
char mp[MX][MX];
char op[5];
char read(){
char c = getchar();
if(c == ' ' || c == '\n'){
c = getchar();
}
return c;
}
int check(){
for(int i = 1; i <= 3; i++){
for(int j = 1; j <= 3; j++){
if(mp[i][j] == op[0]){
if(i+2 <= 3 && mp[i+1][j] == op[0] && mp[i+2][j] == op[0]) return 1;
if(j+2 <= 3 && mp[i][j+1] == op[0] && mp[i][j+2] == op[0]) return 1;
if(i+2 <= 3 && j+2 <= 3 && mp[i+1][j+1] == op[0] && mp[i+2][j+2] == op[0]) return 1;
if(i+2 <= 3 && j-3 >= 0 && mp[i+1][j-i] == op[0] && mp[i+2][j-2] == op[0]) return 1;
}
}
}
return 0;
}
int main(){
int t;
scanf("%d", &t);
while(t--){
for(int i = 1; i <= 3; i++){
for(int j = 1; j <= 3; j++){
mp[i][j] = read();
}
}
scanf("%s", op);
int flag = 0;
for(int i = 1; i <= 3; i++){
for(int j = 1; j <= 3; j++){
if(mp[i][j] == '.'){
int cnt = 0;
mp[i][j] = op[0];
if(check()){
flag = 1;
break;
}
else{
for(int p = 1; p <= 3; p++){
for(int q = 1; q <= 3; q++){
if(mp[p][q] == '.'){
mp[p][q] = op[0];
if(check()){
cnt++;
}
mp[p][q] = '.';
}
}
}
if(cnt >= 2){
flag = 1;
break;
}
}
mp[i][j] = '.';
}
}
}
if(flag) puts("Kim win!");
else puts("Cannot win!");
}
return 0;
}