白书第一章例题6
构造。思维。几何。
分别从几个角度去看,有矛盾就删掉,最后遍历一下统计个数
方法证明:第一个方块肯定要删除。假设前k个必须删除,第k+1个矛盾出现,假如不删掉,矛盾将持续存在,故必须删掉。
代码有很多细节。
比如注意宏定义加() //#define REP(i,n) for(int i=0;i<(n);i++)
#include<iostream>
using namespace std;
#define REP(i,n) for(int i=0;i<(n);i++)
int n;
char read_char() {
char c;
for (;; ) {
c = getchar();
if ((c >= 'A'&&c <= 'Z') || c == '.') return c; }
}void get(int k, int i, int j, int p, int &x, int &y, int &z) {
if (k == ) { x = p; y = j; z = i; }
if (k == ) { x = n - - j; y = p; z = i; }
if (k == ) { x = n - - p; y = n - - j; z = i; }
if (k == ) { x = j; y = n - p - ; z = i; }
if (k == ) { x = n - - i; y = j; z = p; }
if (k == ) { x = i; y = j; z = n - - p; }
}
;
char view[][][], pos[][][];
int main() { while (cin >> n) {
if (n == )break;
REP(i, n) REP(k, ) REP(j, n) view[k][i][j] = read_char();
REP(i, n) REP(j, n) REP(k, n) pos[i][j][k] = '#'; REP(k, ) REP(i, n) REP(j, n) if (view[k][i][j] == '.') { REP(p, n) { int x, y, z; get(k, i, j, p, x, y, z); pos[x][y][z] = '.'; } }; for (;;) { bool done = true;
REP(i, n)REP(j, n)REP(k, ) if (view[k][i][j]!='.' ) {
REP(p, n) {
int x, y, z;
get(k, i, j, p, x, y, z);
if (pos[x][y][z] == '.') continue;
if (pos[x][y][z] == '#') { pos[x][y][z] = view[k][i][j]; break; } if (pos[x][y][z] == view[k][i][j]) break;
pos[x][y][z] = '.';
done = false;
}
}
if (done) break;
}
int ans = ;
REP(i, n)REP(j, n)REP(k, n)
if (pos[i][j][k] != '.') ans++;
printf("Maximum weight: %d gram(s)\n", ans);
}
return ;
}