第一篇博客（CodeForces 754B ）

B. Ilya and tic-tac-toe game

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Ilya is an experienced player in tic-tac-toe on the4 × 4 field. He always starts and plays with Xs. He played a lot of games today with his friend Arseny. The friends became tired and didn't finish the last game. It was Ilya's turn in the game when they left it. Determine whether Ilya could have won the game by making single turn or not.

The rules of tic-tac-toe on the 4 × 4 field are as follows. Before the first turn all the field cells are empty. The two players take turns placing their signs into empty cells (the first player places Xs, the second player places Os). The player who places Xs goes first, the another one goes second. The winner is the player who first gets three of his signs in a row next to each other (horizontal, vertical or diagonal).

The tic-tac-toe position is given in four lines.

Each of these lines contains four characters. Each character is '.' (empty cell), 'x' (lowercase English letterx), or 'o' (lowercase English lettero). It is guaranteed that the position is reachable playing tic-tac-toe, and it is Ilya's turn now (in particular, it means that the game is not finished). It is possible that all the cells are empty, it means that the friends left without making single turn.

PRint single line: "YES" in case Ilya could have won by making single turn, and "NO" otherwise.

xx...oo.x...oox.Output
YESInput
x.oxox..x.o.oo.xOutput
NOInput
x..x..ooo...x.xoOutput
YESInput
o.x.o....x..ooxxOutput
NO
原题链接
题目大意：一个4*4的棋盘，以井字棋的规则，输入棋局现状，如果"x"下一步落子就能获胜就打印"YES"，否则打印"NO"。
分析：
                  最坏落子情况有4*4=16种，每种情况共有12种获胜情形（为端点8种，为中间点有4种），每种获胜的情形需要判断2个位置的情况，所以最坏复杂度为：16*12*4<1000, 因此只需要模拟下棋，将每种情况都考虑在内即可。
下面是AC代码：（题目比较简单但是代码还是写的比较繁琐）
#include<iostream>#include<cstring>#include<math.h>#include<stdlib.h>#include<cstdio>#include<algorithm>using namespace std;char arr[4][4];bool solve(int a, int b){    int dx[] = {-1, -1, -1, 0, 1, 1, 1, 0};    int dy[] = {-1, 0, 1, 1, 1, 0, -1, -1};    for(int i=0; i<8; i++)    {        //这种情形可行（没有越界）        if(a+dx[i]>=0&&a+dx[i]<=3 && b+dy[i]>=0&&b+dy[i]<=3 && a+2*dx[i]>=0&&a+2*dx[i]<=3 && b+2*dy[i]>=0&&b+2*dy[i]<=3)           if(arr[a+dx[i]][b+dy[i]]=='x' && arr[a+2*dx[i]][b+2*dy[i]]=='x')                return true;    }    for(int i=0; i<4; i++)    {        if(a+dx[i]>=0&&a+dx[i]<=3 && b+dy[i]>=0&&b+dy[i]<=3 && a+dx[i+4]>=0&&a+dx[i+4]<=3 && b+dy[i+4]>=0&&b+dy[i+4]<=3)            if(arr[a+dx[i]][b+dy[i]]=='x' && arr[a+dx[i+4]][b+dy[i+4]]=='x')                return true;    }    return false;}int main(void){    //freopen("input.txt","r",stdin);    while(scanf("%s", arr[0])!=EOF)    {        for(int i=0; i<3; i++)            scanf("%s", arr[i+1]);        int flag = 0;//变1表示确定获胜        for(int i=0; i<4; i++)        {            for(int j=0; j<4; j++)            {                if(arr[i][j] == '.')//下一步下在这里                    if(solve(i,j))//确定输赢                    {                        printf("YES/n");                        flag = 1;                        break;                    }            }            if(flag)//确定输赢                break;        }        if(flag == 0)            printf("NO/n");    }    return 0;}如有不妥之处还望各位大佬不吝赐教，在下不胜感激。在此先行谢过。