UVA 12171 (hdu 2771)sculptrue(离散化)

以前对离散化的理解不够,所以把端点和区间区分来考虑但是做完这题以后有了新的认识:

先来看一个问题:给你以下的网格,你需要多少空间去存储红点区间的信息呢?

只需要图上所示的1,2,3,4个点就足够表示红点所在区间了,为什么不是一个区间的第一个红点和最后一个红点呢?(如果这样记录的话则必须加一区间点,记录区间内部信息,因为端点可能是两个区间的交集而区间内可能只被操作了一次)这样做的好处是空白区域的长度也能轻易计算出来。

因此离散化的核心在于以点代表区间

为了计算总区间两端空白的长度,增加A和B点。

原数据找离散后的值直接二分,没必要建map,因为map的查询也是logN级别的

知道了这些,存储问题就解决了。

/*
Created by Rey Chen on 2015.6.25
*/

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
//#define local
using namespace std;
const int maxn = 50 + 1;
const int max2n = 100 + 2;
const int maxc = 1000 + 1;//coordinate;
typedef int ll;
// original data
int n, x0[maxn], y0[maxn], z0[maxn], x1[maxn], y1[maxn], z1[maxn];
// discretized data
int xs[max2n], ys[max2n], zs[max2n];
int xn, yn, zn;//the numbers of each coordinate

//1 means solid, 2 means visited air
int color[max2n][max2n][max2n];


void discretize(int* x, int& n) {
  sort(x, x+n);
  n = unique(x, x+n) - x;
}

struct Cell
{
   int x, y, z;
   Cell(int x,int y,int z):x(x),y(y),z(z){}
};
//related to flood fill
const int dx[] = {1,-1,0,0,0,0};
const int dy[] = {0,0,1,-1,0,0};
const int dz[] = {0,0,0,0,1,-1};

#define vis(u) (color[v.x][v.y][v.z] == 2)
#define setVis(u) color[u.x][u.y][u.z] = 2//solid..
#define uNeigh(i) Cell v(u.x+dx[i], u.y+dy[i], u.z+dz[i])
#define inside(v) (v.x >=0 && v.x < xn && v.y >= 0 && v.y < yn && v.z >= 0 && v.z < zn)
#define issolid(v) (color[v.x][v.y][v.z] == 1)
#define dis(u,x) (x##s[u.x+1]-x##s[u.x])
#define vol(u) (ll)(xs[u.x+1]-xs[u.x])*(ys[u.y+1]-ys[u.y])*(zs[u.z+1]-zs[u.z])//为了让原本最后一层空气计算时不下标不越界 还要再加一层

inline ll area(Cell& v, int dir){
   if(dx[dir]) return (ll)dis(v,y)*dis(v,z);
   if(dy[dir]) return (ll)dis(v,x)*dis(v,z);
   return (ll)dis(v,x)*dis(v,y);
}

//dfs might stack overflow
void floodfill(ll &V,ll &s)
{
   queue<Cell> q;
   Cell u(0,0,0);// Cell u = Cell() 1,2
   setVis(u);
   q.push(u);
   V = s = 0;
   while(!q.empty()) {
      u = q.front();q.pop();
      V += vol(u); //calculate air volume
      for(int i = 0; i < 6; i++) {
         uNeigh(i);
         if(!inside(v)) continue;//最后一层不加
         if(issolid(v)) s += area(v,i); //ensure air vis once meet solid add area
         else if(!vis(v)) {
            setVis(v);
            q.push(v);
         }
      }
   }
   V = maxc*maxc*maxc - V;
}

#define add(x) x##1[i] += x##0[i]
#define dadd(a) xs[t] = x##a[i]; ys[t] = y##a[i]; zs[t++] = z##a[i]
#define ID(x,n,val) (lower_bound(x,x+n,val) - x)
#define GetId(X,x) X##0 = ID(x##s, x##n, x##0[i]);X##1 = ID(x##s, x##n, x##1[i])
//X0 = ID(xs,xn,x0[i]);
int main()
{
#ifdef local
    freopen("in.txt","r",stdin);
#endif // local

   int T;
   scanf("%d",&T);
   while(T--) {
      scanf("%d",&n); int t = 2;
      *xs = *ys = *zs = 0;//add air 
      xs[1] = ys[1] = zs[1] = maxc;//最后面有两层空气
      for(int i = 0; i < n; i++) {
         scanf("%d%d%d%d%d%d", x0+i, y0+i, z0+i, x1+i, y1+i, z1+i);
         add(x);add(y);add(z);
         dadd(0); dadd(1);
      }
      xn = yn = zn = t ;
      discretize(xs,xn);
      discretize(ys,yn);
      discretize(zs,zn);
      xn--;yn--;zn--;//不访问最后一层空气,计算总体积时要加上最后一层空气

      memset(color,0,sizeof(color));
      for(int i = 0; i < n; i++) { //solid set
         int X0, X1, Y0, Y1, Z0, Z1;
         GetId(X,x); GetId(Y,y); GetId(Z,z);
         for(int X = X0; X < X1; X++)
            for(int Y = Y0; Y < Y1; Y++)
               for(int Z = Z0; Z < Z1; Z++)
                  color[X][Y][Z] = 1;
      }
      ll v,s;
      floodfill(v,s);
      printf("%d %d\n",s,v);
   }
   return 0;
}

 

posted @ 2015-06-25 10:42  陈瑞宇  阅读(1246)  评论(0编辑  收藏  举报