Uva LV 2995 Image Is Everything 模拟,坐标映射,视图映射 难度: 1
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题目
https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=996
题意
被射击掉了一些列的魔方。组成模仿的每个方块的六个面颜色相同。给六个面的视图,求魔方最多还剩下多少个小块。
魔方最多十阶。
思路
关键在于将视图的坐标(加上视图深度)映射为三维坐标系内的坐标,之后就可以不断删除会造成矛盾的暴露在表面的方块,直到没有方块或者没有矛盾为止。
感想
1. 注意<写成了>=
2. 忘记了检测当前方块是否已经删除过了。
3. 忘了memset
4. 视图的映射有价值,不过需要注意坐标系建立不同
代码
Runtime: 0.006
#include <algorithm> #include <cassert> #include <cmath> #include <cstdio> #include <cstring> #include <iostream> #include <queue> #include <tuple> #include <cassert> using namespace std; #define LOCAL_DEBUG const int MAXN = 11; int n; bool del[MAXN][MAXN][MAXN]; char color[MAXN][MAXN][MAXN]; char fc[6][MAXN][MAXN]; int deep[6][MAXN][MAXN]; int facec[3]; int polarc[3]; char statusDescribe[6][4] = { "YXZ", "ZXy", "yXz", "zXY", "YZx", "YzX" }; int getAns() { int ans = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { for (int k = 0; k < n; k++) { if (!del[i][j][k])ans++; } } } return ans; } void face2polar(int fid, int facec[3], int polarc[3]) { for (int i = 0; i < 3; i++) { char sta = statusDescribe[fid][i]; if (sta >= ‘X‘ && sta <= ‘Z‘) { polarc[i] = facec[sta - ‘X‘]; } else { polarc[i] = n - 1 - facec[sta - ‘x‘]; } } } void polar2face(int fid, int facec[3], int polarc[3]) { for (int i = 0; i < 3; i++) { char sta = statusDescribe[fid][i]; if (sta >= ‘X‘ && sta <= ‘Z‘) { facec[sta - ‘X‘] = polarc[i]; } else { facec[sta - ‘x‘] = n - 1 - polarc[i]; } } } int main() { #ifdef LOCAL_DEBUG freopen("input.txt", "r", stdin); //freopen("output2.txt", "w", stdout); #endif // LOCAL_DEBUG for (int ti = 1; scanf("%d", &n) == 1 && n; ti++) { memset(del, 0, sizeof(del)); memset(color, 0, sizeof(color)); memset(fc, 0, sizeof(fc)); memset(deep, 0, sizeof(deep)); memset(facec, 0, sizeof(facec)); memset(polarc, 0, sizeof(polarc)); char buff[100]; cin.getline(buff, 100); for(int i = 0; i < n;i++){ cin.getline(buff, 100); for (int k = 0; k < 6; k++) { for (int j = 0; j < n; j++) { fc[k][i][j] = buff[(n + 1) * k + j]; } } } int &x = facec[0]; int &y = facec[1]; int &z = facec[2]; int &a = polarc[0]; int &b = polarc[1]; int &c = polarc[2]; for (int fid = 0; fid < 6; fid++) { for (x = 0; x < n; x++) { for (y = 0; y < n; y++) { if (fc[fid][x][y] == ‘.‘) { deep[fid][x][y] = n; for (z = 0; z < n; z++) { face2polar(fid, facec, polarc); // printf("R: %d, del %d-(%d, %d, %d) : (%d, %d, %d) ", getAns(), fid, x, y, z, a, b, c); del[a][b][c] = true; } } } } } bool checkFlag = true; while (checkFlag) { checkFlag = false; for (int fid = 0; fid < 6; fid++) { for (x = 0; x < n; x++) { for (y = 0; y < n; y++) { for (z = deep[fid][x][y]; z < n; z++, deep[fid][x][y]++) { face2polar(fid, facec, polarc); if (del[a][b][c])continue; else if (color[a][b][c] == 0 || color[a][b][c] == fc[fid][x][y]) { color[a][b][c] = fc[fid][x][y]; break; } else{ del[a][b][c] = true; checkFlag = true; } } } } } } int ans = getAns(); printf("Maximum weight: %d gram(s) ", ans); } return 0; }
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