[Codeforces]1015E1. Stars Drawing (Easy Edition)

題目URL:http://codeforces.com/contest/1015/problem/E1
淹水淹水淹水法啦~~~~
其實可以建四個方向前綴，取size的min，再O(n^2)比較，但O(n^3)可以過的話有何不可呢??(只可惜E2的hard edition就QQ了~)

#include<bits/stdc++.h>
using namespace std;
char m[1005][1005];
int sor[4],a,b,total=0,tot=0,cou=0,flag=0,f=0;

int ans[10005],ansi[10005],ansj[10005],stri[10005],strj[10005];
signed main()
{
    cin >> a >> b;
    for(int i=1;i<=a;i++)
    for(int j=1;j<=b;j++)
    {
        cin >> m[i][j];

        if(m[i][j]=='*')

        {

            stri[flag]=i;

            strj[flag]=j;

            flag++;

        }

    }
    int aa=0,bb=0,cc=0,dd=0;
    for(int i=0;i<flag;i++)
    {
        aa=0;

        tot=1;

        while((m[stri[i]][strj[i]-tot]=='*' || m[stri[i]][strj[i]-tot]=='#')&& \

        (m[stri[i]][strj[i]+tot]=='*' || m[stri[i]][strj[i]+tot]=='#')&& \

        (m[stri[i]+tot][strj[i]]=='*' || m[stri[i]+tot][strj[i]]=='#')&& \

        (m[stri[i]-tot][strj[i]]=='*' || m[stri[i]-tot][strj[i]]=='#'))

        {

            m[stri[i]+tot][strj[i]]='#';

            m[stri[i]-tot][strj[i]]='#';

            m[stri[i]][strj[i]+tot]='#';

            m[stri[i]][strj[i]-tot]='#';

            m[stri[i]][strj[i]]='#';

            tot++;

            aa++;

        }

        ansi[f]=stri[i];

        ansj[f]=strj[i];

        ans[f]=aa;

        f++;

    }
    int deter=0;
    for(int i=1;i<=a;i++)
    {
        for(int j=1;j<=b;j++)

        {

            //cout << m[i][j];

            if(m[i][j]=='*')

            {

                deter=1;

            }

        }

        //cout << endl;

    }
    int co=0;
    if(deter) return cout << -1 ,0;
    for(int i=0;i<f;i++)
    {
        if(ans[i]!=0)

        co++;

    }

    cout << co << endl;
    for(int i=0;i<f;i++)
    {
        if(ans[i]!=0)

        cout << ansi[i] << " " << ansj[i] << " " << ans[i] << endl;

    }
    return 0;
}