【习题 4-2 Uva201】Squares

【链接】 我是链接,点我呀:)

【题意】

在这里输入题意

【题解】

注意那个星号的数量。。。
然后V x y的话,是从(y,x)向(y+1,x)连线。
H x y才是从(x,y)向(x,y+1)连线
枚举以(x,y)作为左上角的举行就ok了

【代码】

#include <bits/stdc++.h>

using namespace std;

const int N = 10;

int n,m;

bool a[N+10][N+10][2];

int cnt[N+5];

bool check(int x,int y,int p){

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

        if (a[x][y][0]){

            y++;

            if (y>n) break;

        }else return 0;

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

        if (a[x][y][1]){

            x++;

            if (x>n) break;

        }else return 0;

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

        if (y-1>=1 && a[x][y-1][0]){

            y--;

            if (y<1) break;

        }else return 0;

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

        if (x-1>=1 && a[x-1][y][1]){

            x--;

            if (x<1) break;

        }else return 0;

    return 1;

}

int main(){

    //freopen("/home/ccy/rush.txt","r",stdin);

    //freopen("/home/ccy/rush_out.txt","w",stdout);

    ios::sync_with_stdio(0),cin.tie(0);

    int kase = 0;

    while (cin >> n){

        if (kase>0){

            cout<<endl;

            cout<<"**********************************"<<endl;

            cout<<endl;

        }

        cout<<"Problem #"<<++kase<<endl<<endl;

        memset(a,0,sizeof a);

        memset(cnt,0,sizeof cnt);

        cin >> m;

        for (int i = 1;i <= m;i++){

            char s[5];int x,y;

            cin >> s >> x >> y;

            if (s[0]=='H')

                a[x][y][0] = 1;

            else{

                swap(x,y);

                a[x][y][1] = 1;

            }

        }

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

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

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

                    if (check(i,j,p))

                        cnt[p]++;

        bool none = 1;

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

            if (cnt[i]>0) {

                none = 0;

                cout<<cnt[i]<<" square (s) of size "<<i<<endl;

            }

        }

        if (none) cout<<"No completed squares can be found."<<endl;

    }

    return 0;

}