Educational Codeforces Round 25 B Five-In-a-Row 基础题
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CF传送门
题解:问在棋盘的空位上再放一个X棋是否能使X胜利
题解:
1. 暴力搜索就行了,图才10X10,随便嵌套五六个循环都不会超时,所以这题就随便写
2. CF给出的题解是直接对图进行遍历看看是否存在五个连线点上是4个"X"和1个"."的情况,有就YES
3. 我写的是先遍历图把可能的点补上再搜五连点,我的复杂度要比CF题解多出几个循环,但还是过了
以下是我的AC代码:
#include <iostream>using namespace std;char s[15][15];bool win(){ for(int i=0;i<10;i++) { for(int j=0;j<10;j++) { if(s[i][j]=='X') { int u=i,v=j,sum=0; while(v<10&&s[u][v]=='X'){sum++;v++;} //右直线 if(sum==5) return true; u=i;v=j;sum=0; while(u<10&&s[u][v]=='X'){sum++;u++;} //下直线 if(sum==5) return true; u=i;v=j;sum=0; while(u<10&&v<10&&s[u][v]=='X'){sum++;u++;v++;} //右下直线 if(sum==5) return true; u=i;v=j;sum=0; while(u<10&&v>=0&&s[u][v]=='X'){sum++;u++;v--;} //左下直线 if(sum==5) return true; } } } return false;}int main(){ for(int i=0;i<10;i++) cin >> s[i]; for(int i=0;i<10;i++) for(int j=0;j<10;j++) if(s[i][j]=='X') for(int v=-1;v<=1;v++) for(int u=-1;u<=1;u++) if(i+v>=0&&i+v<10&&j+u>=0&&j+u<10&&s[i+v][j+u]=='.'){ s[i+v][j+u]='X'; //更改点的状态 if(win()){ cout << "YES" << endl; return 0; } s[i+v][j+u]='.'; //记得还原改变的点的状态 } cout << "NO" << endl; return 0;}
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