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  • CodeForces 1059B Forgery(模拟)

    题目链接:https://vjudge.net/problem/CodeForces-1059B

    题目大意:你有一个可以盖出一个3*3 ‘#’中心为‘.’的章,问你能不能盖出图中的形状

      其实简单模拟一下就行了.....如果盖下一个章,那么这个章的中心8个方向一定是'#',那我们反过来做一个'.'图,只要原图某个点的8个方向都是'#',那我们就以这个点为中心盖一个章,最后对比一下我们构造的图和原图,如果不一样,那肯定是无法盖出图中的形状的.

    #include<set>
    #include<map>
    #include<stack>
    #include<queue>
    #include<cmath>
    #include<cstdio>
    #include<cctype>
    #include<string>
    #include<vector>
    #include<climits>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    #define endl '\n'
    #define rtl rt<<1
    #define rtr rt<<1|1
    #define lson rt<<1, l, mid
    #define rson rt<<1|1, mid+1, r
    #define maxx(a, b) (a > b ? a : b)
    #define minn(a, b) (a < b ? a : b)
    #define zero(a) memset(a, 0, sizeof(a))
    #define INF(a) memset(a, 0x3f, sizeof(a))
    #define IOS ios::sync_with_stdio(false)
    #define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n")
    using namespace std;
    typedef long long ll;
    typedef pair<int, int> P;
    typedef pair<ll, ll> P2;
    const double pi = acos(-1.0);
    const double eps = 1e-7;
    const ll MOD =  1000000007LL;
    const int INF = 0x3f3f3f3f;
    const int _NAN = -0x3f3f3f3f;
    const double EULC = 0.5772156649015328;
    const int NIL = -1;
    template<typename T> void read(T &x){
        x = 0;char ch = getchar();ll f = 1;
        while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
        while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
    }
    const int maxn = 1e3+10;
    char omap[maxn][maxn], vmap[maxn][maxn];
    int dir[8][2] = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};
    bool ok(int x, int y) {
        for (int i = 0; i<8; ++i) {
            int newx = x+dir[i][0], newy = y+dir[i][1];
            if (omap[newx][newy]=='.') return false;
        }
        return true;
    }
    void make(int x, int y) {
        for (int i = 0; i<8; ++i) {
            int newx = x+dir[i][0], newy = y+dir[i][1];
            vmap[newx][newy] = '#';
        }
    }
    bool checker(int n, int m) {
        for (int i = 0; i<n; ++i)
            for (int j = 0; j<m; ++j)
                if (omap[i][j]!=vmap[i][j]) return false;
        return true;
    }
    int main(void) {
        int n, m;
        while(~scanf("%d%d", &n, &m)) {
            for (int i = 0; i<n; ++i)
                scanf("%s", omap[i]);
            for (int i = 0; i<n; ++i) //全部初始化为.图
                for (int j = 0; j<m; ++j)
                    vmap[i][j] = '.';
            for (int i = 1; i<n-1; ++i) 
                for (int j = 1; j<m-1; ++j) //参考样例2,中心不能在边缘
                    if (ok(i, j)) make(i, j); //如果一个点8个方向都有#那么就以这个点为中心盖一个章
            printf(checker(n, m) ? "YES\n" : "NO\n"); //对比两个图是否一样
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/shuitiangong/p/12401585.html
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