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

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

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了。。。
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]
= '\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;
}