Pawn‘s Revenge (暴力)

先将k周围标记，然后找出所有失败的情况。剩下的就是一定会成功的情况，对于每个‘*’，先判断左下是否已经放过兵，如果没有，在它右下放兵并标记，结果+1。如果它的右下无法放，在左下放兵并标记，结果+1。

#include <stdio.h>
#include <iostream>
using namespace std;
char bor[1005][1005];
int vis[1005][1005]={0};
int pass[1005][1005]={0};
int ans[1005][1005]={0};
int main()
{int n;int sum=0;cin >> n;getchar();for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){cin >> bor[i][j];if(bor[i][j]=='K'){pass[i-1][j-1]=1;pass[i-1][j]=1;pass[i-1][j+1]=1;pass[i][j-1]=1;pass[i][j+1]=1;pass[i+1][j-1]=1;pass[i+1][j]=1;pass[i+1][j+1]=1;}}getchar();}int flag=0;for(int j=1;j<=n;j++){if(bor[n][j]=='*'&&pass[n][j]==0){flag=1;break;}}for(int i=1;i<n;i++){for(int j=1;j<=n;j++){if(bor[i][j]=='*'&&pass[i][j]==0&&bor[i+1][j-1]!='w'){if((j==1&&bor[i+1][j+1]=='*')||(j==n&&bor[i+1][j-1]=='*')){flag=1;break;}else if(bor[i+1][j-1]=='*'&&bor[i+1][j+1]=='*'){flag=1;break;}else{if(bor[i+1][j+1]=='-'){sum++;bor[i+1][j+1]='w';}else if(bor[i+1][j-1]=='-'){sum++;bor[i+1][j-1]='w';}}}}}if(flag==1){cout << -1;}else{cout << sum;}}