uva806 Spatial Structures 空间结构 （黑白图像的四分树表示）

 input

8

00000000

00000000

00001111

00001111

00011111

00111111

00111100

00111000

-8

9 14 17 22 23 44 63 69 88 94 113 -1

2

00

00

-4

0 -1

0

output

Image 1

9 14 17 22 23 44 63 69 88 94 113

Total number of black nodes = 11



Image 2

........

........

....****

....****

...*****

..******

..****..

..***...



Image 3

Total number of black nodes = 0



Image 4

****

****

****

****

这题目就是给你黑白点图的两种表示方法让你互相转化，写的时候没注意格式。
数字是代表四分树黑色结点到根的距离，由于定义是递归的，所以不必真的把树建出来。

/*

Created by Rey Chen on 2015.6.30

Copyright (c) 2015 Rey . All rights reserved.

*/



#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 65;



char G[maxn][maxn];



int path[maxn*maxn];

int sz;



void solve1(int r1,int c1,int r2,int c2,int wei,int sum)

{

   bool whi, bla; whi = bla = false;

   for(int i = r1;i <= r2;i++)

      for(int j = c1;j <= c2;j++){

         if(!bla && G[i][j] == '1') bla = true;

         if(!whi && G[i][j] == '0') whi = true;

         if(whi&&bla) { break;}

      }

   if(bla) {

      if(!whi) {path[++sz] = sum; return;}

   }else if(whi) return;



   int dvr = r1+r2>>1, dvc = c1+c2>>1;

      int nwei = 5*wei;

   solve1(   r1,   c1,dvr,dvc,nwei,sum+ wei    );

   solve1(   r1,dvc+1,dvr, c2,nwei,sum+(wei<<1));

   solve1(dvr+1,   c1 ,r2,dvc,nwei,sum+  wei*3 );

   solve1(dvr+1,dvc+1, r2 ,c2,nwei,sum+(wei<<2));

}



int a[20];

void draw(int n,int r1,int c1,int r2,int c2){

   int Sz = 0;

   while(n){ a[Sz++]=n%5;n/=5;}

   for(int i = 0; i < Sz;i++){

      switch (a[i]){

         case 1:{

            r2 = r1+r2>>1; c2 = c1+c2>>1;

            break;

         }

         case 2:{

            r2 = r1+r2>>1; c1 = (c1+c2>>1)+1;

            break;

         }

         case 3:{

            r1 = (r1+r2>>1)+1; c2 = c1+c2>>1;

            break;

         }

         case 4:{

            r1 = (r1+r2>>1)+1; c1 = (c1+c2>>1)+1;

            break;

         }

      }

   }

   for(int i = r1;i <= r2;i++)

      for(int j = c1;j <= c2;j++)

         G[i][j] = '*';

}





int main()

{

   int n;

   int Cas = 0;

   while(~scanf("%d",&n)&&n){

      if(Cas) puts("");

      if(n>0){

         for(int i = 0; i < n; i++)

            scanf("%s",G[i]);

         sz = 0;

         solve1(0,0,n-1,n-1,1,0);

         sort(path+1,path+sz+1);



         printf("Image %d\n",++Cas);

         for(int i = 1; i < sz; i++)//PE ,bi dog

            printf("%d%c",path[i],i%12?' ':'\n');

         if(sz)printf("%d\n",path[sz]);

         printf("Total number of black nodes = %d\n",sz);

      }else {

         n = -n;

         int t;

         for(int i = 0;i < n;i++){

            for(int j = 0;j < n;j++)

               G[i][j]='.';

            G[i][n] = '\0';

         }



         while(scanf("%d",&t)&&~t){

            draw(t,0,0,n-1,n-1);

         }

         printf("Image %d\n",++Cas);

         for(int i = 0;i < n;i++)

            puts(G[i]);

      }



   }

   return 0;

}