UVa1030 Image Is Everything

思路：针对每个视图，列举出其可能处在的空间坐标，如果两次计算得到的颜色不一致，说明该点不存在。直到没有这种情况出现为止，另外这种计算的是最大值情况。

代码如下：

#include <iostream>
#include <cstdio>
#include <fstream>using namespace std;const int N = 11;
const int SIX = 6;#define REP(i, n) for (int i = 0; i < n; ++i)char view[SIX][N][N];
char pos[N][N][N];char readChar()
{char ch;while (1){ch = cin.get();if ((('A' <= ch) && ('Z' >= ch)) || ('.' == ch)){return ch;}}
}void getPos(int n, int v, int i, int j, int len, int& x, int& y, int& z)
{if (0 == v){x = len; y = j; z = i;}else if (1 == v){x = n - 1 - j; y = len; z = i;}else if (2 == v){x = n - 1 - len; y = n - 1 - j; z = i;}else if (3 == v){x = j; y = n - 1 - len; z = i;}else if (4 == v){x = n - 1 - i; y = j; z = len;} else{x = i; y = j; z = n - 1 - len;}
}int solve(int n)
{REP(i, n) REP(j, SIX) REP(k, n) view[j][i][k] = readChar();REP(i, n) REP(j, n) REP(k, n) pos[i][j][k] = '#';REP(i, SIX) REP(j, n) REP(k, n) if ('.' == view[i][j][k]){REP(p, n){int x, y, z;getPos(n, i, j, k, p, x, y, z);pos[x][y][z] = '.';}}for (;;){bool done = true;REP(i, SIX) REP(j, n) REP(k, n){if ('.' != view[i][j][k]){REP(p, n){int x, y, z;getPos(n, i, j, k, p, x, y, z);if ('.' == pos[x][y][z]) continue;if ('#' == pos[x][y][z]){pos[x][y][z] = view[i][j][k];break;}if (pos[x][y][z] == view[i][j][k]) break;pos[x][y][z] = '.';done = false;}}}if (done) break;}int ans = 0;REP(i, n) REP(j, n) REP(k, n){if ('.' != pos[i][j][k]) ++ans;}return ans;
}int main()
{
#ifndef ONLINE_JUDGEifstream fin("f:\\OJ\\uva_in.txt");streambuf *back = cin.rdbuf(fin.rdbuf());
#endifint n;while (cin >> n, n){int ans = solve(n);cout << "Maximum weight: " << ans << " gram(s)" << endl;}#ifndef ONLINE_JUDGEcin.rdbuf(back);
#endifreturn 0;
}

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