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UVa 12171 (离散化 floodfill) Sculpture

题意:

三维空间中有n个长方体组成的雕塑,求表面积和体积。

分析:

我们可以在最外边加一圈“空气”,然后求空气的连通块的体积,最后用总体积减去即是雕塑的体积。

还有一个很“严重”的问题就是5003所占的空间太大,因此需要离散化。而在计算体积和表面积的时候要用原坐标。

离散化以后的坐标分别保存在xs、ys、zs,坐标为(x, y, z)的格子代表([xs[x], ys[y], zs[z]) ~ (xs[x+1], ys[y+1], zs[z+1]) 这一个小长方体。

这个题的难度对我来说属于大概思路比较明白,但是很多代码细节处理不好那种。

把节点和相关的函数封装在一个结构体里面是个狠不错的技巧,使编码思路清晰,代码可读性也很好。

技术分享
  1 #include <cstdio>  2 #include <algorithm>  3 #include <queue>  4 #include <cstring>  5 using namespace std;  6   7 const int maxn = 50 + 5;  8 const int maxc = 1000 + 1;  9  10 int n, x0[maxn], y0[maxn], z0[maxn], x1[maxn], y1[maxn], z1[maxn]; 11  12 int nx, ny, nz; 13 int xs[maxn*2], ys[maxn*2], zs[maxn*2]; 14  15 const int dx[] = {1,-1,0,0,0,0}; 16 const int dy[] = {0,0,1,-1,0,0}; 17 const int dz[] = {0,0,0,0,1,-1}; 18 int color[maxn*2][maxn*2][maxn*2]; 19  20 struct Cell 21 { 22     int x, y, z; 23     Cell(int x=0, int y=0, int z=0):x(x), y(y), z(z) {} 24     bool valid() const { return x >= 0 && x < nx-1 && y >= 0 && y < ny-1 && z >= 0 && z < nz-1;} 25     bool solid() const { return color[x][y][z] == 1; } 26     bool getVis() const { return color[x][y][z] == 2; } 27     void setVis() const { color[x][y][z] = 2; } 28     Cell neighbor(int dir) const 29     { return Cell(x+dx[dir], y+dy[dir], z+dz[dir]); } 30     int volume() 31     { return (xs[x+1]-xs[x]) * (ys[y+1]-ys[y]) * (zs[z+1]-zs[z]); } 32     int area(int dir) 33     { 34         if(dx[dir]) return (ys[y+1]-ys[y]) * (zs[z+1]-zs[z]); 35         if(dy[dir]) return (xs[x+1]-xs[x]) * (zs[z+1]-zs[z]); 36         return (xs[x+1]-xs[x]) * (ys[y+1]-ys[y]); 37     } 38 }; 39  40 void discrectize(int* x, int& n) 41 { 42     sort(x, x + n); 43     n = unique(x, x + n) - x; 44 } 45  46 int ID(int* x, int n, int x0) 47 { 48     return lower_bound(x, x + n, x0) - x; 49 } 50  51 void floodfill(int& v, int& s) 52 { 53     v = s = 0; 54     Cell c; 55     c.setVis(); 56     queue<Cell> q; 57     q.push(c); 58     while(!q.empty()) 59     { 60         Cell c = q.front(); q.pop(); 61         v += c.volume(); 62         for(int i = 0; i < 6; ++i) 63         { 64             Cell c2 = c.neighbor(i); 65             if(!c2.valid()) continue; 66             if(c2.solid()) s += c.area(i); 67             else if(!c2.getVis()) 68             { 69                 c2.setVis(); 70                 q.push(c2); 71             } 72         } 73     } 74     v = maxc*maxc*maxc - v; 75 } 76  77 int main() 78 { 79     //freopen("in.txt", "r", stdin); 80     int T; 81     scanf("%d", &T); 82     while(T--) 83     { 84         memset(color, 0, sizeof(color)); 85         nx = ny = nz = 2; 86         xs[0] = ys[0] = zs[0] = 0; 87         xs[1] = ys[1] = zs[1] = maxc; 88         scanf("%d", &n); 89         for(int i = 0; i < n; ++i) 90         { 91             scanf("%d%d%d%d%d%d", &x0[i], &y0[i], &z0[i], &x1[i], &y1[i], &z1[i]); 92             x1[i] += x0[i]; y1[i] += y0[i]; z1[i] += z0[i]; 93             xs[nx++] = x0[i]; xs[nx++] = x1[i]; 94             ys[ny++] = y0[i]; ys[ny++] = y1[i]; 95             zs[nz++] = z0[i]; zs[nz++] = z1[i]; 96         } 97         discrectize(xs, nx); 98         discrectize(ys, ny); 99         discrectize(zs, nz);100 101         for(int i = 0; i < n; ++i)102         {103             int X1 = ID(xs, nx, x0[i]), X2 = ID(xs, nx, x1[i]);104             int Y1 = ID(ys, ny, y0[i]), Y2 = ID(ys, ny, y1[i]);105             int Z1 = ID(zs, nz, z0[i]), Z2 = ID(zs, nz, z1[i]);106             for(int X = X1; X < X2; X++)107                 for(int Y = Y1; Y < Y2; ++Y)108                     for(int Z = Z1; Z < Z2; ++Z)109                         color[X][Y][Z] = 1;110         }111 112         int v, s;113         floodfill(v, s);114         printf("%d %d\n", s, v);115     }116 117     return 0;118 }
代码君

 

UVa 12171 (离散化 floodfill) Sculpture