题目链接
https://www.nowcoder.com/acm/contest/85/F
思路 记录每一个面 上的点 是否有方块
然后 根据它的输出顺序 遍历访问 如果有 输出 ‘X’ 否则 输出‘.’
但是要注意它的坐标系不是常规的坐标系
AC代码
#include <cstdio> #include <cstring> #include <ctype.h> #include <cstdlib> #include <cmath> #include <climits> #include <ctime> #include <iostream> #include <algorithm> #include <deque> #include <vector> #include <queue> #include <string> #include <map> #include <stack> #include <set> #include <numeric> #include <sstream> #include <iomanip> #include <limits> #define CLR(a) memset(a, 0, sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair <int, int> pii; typedef pair <ll, ll> pll; const double PI = 3.14159265358979323846264338327; const double E = exp(1); const double eps = 1e-6; const int INF = 0x3f3f3f3f; const int maxn = 1e3 + 5; const int MOD = 1e9 + 7; int xoy[maxn][maxn], xoz[maxn][maxn], yoz[maxn][maxn]; int main() { CLR(xoy); CLR(xoz); CLR(yoz); int x, y, z, n; scanf("%d%d%d%d", &x, &y, &z, &n); for (int i = 0; i < n; i++) { int a, b, c; scanf("%d%d%d", &a, &b, &c); xoy[a][b] = 1; xoz[a][c] = 1; yoz[b][c] = 1; } for (int i = y; i >= 1; i--) { for (int j = 1; j <= x; j++) { if (xoy[j][i]) printf("x"); else printf("."); } printf(" "); for (int j = 1; j <= z; j++) { if (yoz[i][j]) printf("x"); else printf("."); } printf("\n"); } printf("\n"); for (int i = 1; i <= z; i++) { for (int j = 1; j <= x; j++) { if (xoz[j][i]) printf("x"); else printf("."); } printf("\n"); } }转载于:https://www.cnblogs.com/Dup4/p/9433238.html
