1250. Sea Burial

http://acm.timus.ru/problem.aspx?space=1&num=1250

最外层加上一层‘#’    然后将island 和 sea  一层层的分离  如果某一层向外接触的 fragment 全是被诅咒的 那么这种被诅咒是会传递到这一层的

代码：

#include<iostream>

#include<cstdio>

#include<cstring>

#include<string>

#include<map>

#include<vector>

#include<stack>

#include<set>

#include<map>

#include<queue>

#include<algorithm>

#include<cmath>

#define LL long long

#define sint short int

//#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const int N=505;

char s[N][N];

bool visited[N][N];

int sacred[N][N];

int X[]={0,0,-1,1,-1,-1,1,1};

int Y[]={-1,1,0,0,-1,1,-1,1};

int n,m,sx,sy;

int ans;

queue<int>qtx,qty;

stack<int>stx,sty;

void bfs(int x1,int y1)

{

    int I=8;

    if(s[x1][y1]=='#')

    I=4;

    int flag = 1;

    if(x1==0&&y1==0)

    flag=-1;

    qtx.push(x1);qty.push(y1);

    stx.push(x1);sty.push(y1);

    visited[x1][y1]=true;

    while(!qtx.empty())

    {

        int x=qtx.front();qtx.pop();

        int y=qty.front();qty.pop();

        for(int i=0;i<I;++i)

        {

            int l1=x+X[i];

            int l2=y+Y[i];

            if(l1>=0&&l1<=n+1&&l2>=0&&l2<=m+1)

            {

                if(visited[l1][l2]&&sacred[l1][l2]==-1)

                flag=-1;

                if(visited[l1][l2]||s[l1][l2]!=s[x][y])

                continue;

                visited[l1][l2]=true;

                qtx.push(l1);

                qty.push(l2);

                stx.push(l1);

                sty.push(l2);

            }

        }

    }

    if(flag==1&&I==4)

    ++ans;

    if(x1==sx&&y1==sy)

    flag=1;

    while(!stx.empty())

    {

       int l1=stx.top();stx.pop();

       int l2=sty.top();sty.pop();

       sacred[l1][l2]=flag;

    }

}

int main()

{

    //freopen("data.in","r",stdin);

    while(cin>>m>>n>>sy>>sx)

    {

        getchar();

        for(int i=1;i<=n;++i)

        gets(s[i]+1);

        memset(visited,false,sizeof(visited));

        memset(sacred,0,sizeof(sacred));

        ans=0;

        for(int j=0;j<=m+1;++j)

        s[0][j]=s[n+1][j]='#';

        for(int i=0;i<=n+1;++i)

        s[i][0]=s[i][m+1]='#';

        bfs(0,0);

        bfs(sx,sy);

        for(int i=1;i<=n;++i)

        for(int j=1;j<=m;++j)

        if(!visited[i][j])

        bfs(i,j);

        cout<<ans<<endl;

    }

    return 0;

}