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;
	}

}