hdu 3682(To Be an Dream Architect)

把座標映射成一個  n進制數  hash一下，去掉重複的即可。處理方法有很多種。    用STL的set、和map會超時

 

val = x*n*n + y *n + z;

 

我用了sort排序然後手動統計不同數目就過了

 

代碼：

 

/* * File: main.cpp * Author: Hewie * * Created on November 15, 2010, 10:06 AM */ #include <cstdlib> #include <cstdio> #include <algorithm> using namespace std; #define MAXN 1001 int num[MAXN*MAXN]; int main(int argc, char** argv) { int T; scanf("%d", &T); while (T--) { int n, m; scanf("%d%d", &n, &m); getchar(); char a, b; int valA, valB; int len = 0, i; while (m--) { scanf("%c=%d,%c=%d", &a, &valA, &b, &valB); getchar(); // printf("%c=%d,%c=%d/n", a, valA, b, valB); if (a == 'X') { if (b == 'Y') for (i = 1; i <= n; ++i) num[len++] = valA * n * n + valB * n + i; else for (i = 1; i <= n; ++i) num[len++] = valA * n * n + i * n + valB; } else if (a == 'Y') { if (b == 'X') for (i = 1; i <= n; ++i) num[len++] = valB * n * n + valA * n + i; else for (i = 1; i <= n; ++i) num[len++] = i * n * n + valA * n + valB; } else { if (b == 'X') for (i = 1; i <= n; ++i) num[len++] = valB * n * n + i * n + valA; else for (i = 1; i <= n; ++i) num[len++] = i * n * n + valB * n + valA; } } sort(num, num + len); int sum(1); int p = 0; for (i = 1; i < len; ++i) { if (num[i] != num[p]) p = i, ++sum; } printf("%d/n", sum); } return 0; }