Problem UVA806-Spatial Structures
Accept：329 Submit：2778
Time Limit: 3000 mSec
Problem Description
Problem Description
Input
The input contains one or more images. Each image is square, and the data for an image starts with an integer n, where |n| is the length of a side of the square (always a power of two, with |n| < 64) followed by a representation of the image. A representation is either a sequence of n2 zeros and ones comprised of |n| lines of |n| digits per line, or the sequence of numbers that represent the root-to-leaf paths of each black node in the quadtree that represents the image. If n is positive, the zero/one representation follows; if n is negative, the sequence of black node path numbers (in base 10) follows. The sequence is terminated by the number -1. A one-node tree that represents an all-black image is represented by the number 0. A one-node tree that represents an all-white image is represented by an empty sequence (no numbers). The end of data is signaled by a value of 0 for n.
Output
For each image in the input, ?rst output the number of the image, as shown in the sample output. Then output the alternate form of the image. If the image is represented by zeros and ones, the output consists of root-to-leaf paths of all black nodes in the quadtree that represents the image. The values should be base 10 representations of the base 5 path numbers, and the values should be printed in sorted order. If there are more than 12 black nodes, print a newline after every 12 nodes. The total number of black nodes should be printed after the path numbers. If the image is represented by the root-to-leaf paths of black nodes, the output consists of an ASCII representation of the image with the character ‘.’ used for white/zero and the character ‘*’ used for black/one. There should be n characters per line for an n×n image.
Sample 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
Sample Ouput
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
****
****
****
****
题解：这个题如果没有lrj前面例题的铺垫，我自己估计是搞不定，不过做了那个例题之后，这个题就是稍微麻烦一点，没什么特殊的地方，重在代码基本功。
这个题的输出格式有点坑，总的来说就是题中没说的换行不要有，尤其是最后一个Case。
四分树例题：UVA297：Quadtrees
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cstdlib>
5 #include <string>
6 #include <vector>
7 #include <algorithm>
8 using namespace std;
9
10 const int maxn = 70;
11 int n,cnt;
12 char gra[maxn][maxn];
13 vector<int> ans;
14
15 bool is_black(int r,int c,int wide){
16 for(int i = r;i < r+wide;i++){
17 for(int j = c;j < c+wide;j++){
18 if(gra[i][j] == ‘0‘) return false;
19 }
20 }
21 return true;
22 }
23
24 bool is_white(int r,int c,int wide){
25 for(int i = r;i < r+wide;i++){
26 for(int j = c;j < c+wide;j++){
27 if(gra[i][j] == ‘1‘) return false;
28 }
29 }
30 return true;
31 }
32
33 int consumption(string &ss){
34 int res = 0;
35 for(int i = ss.size()-1;i >= 0;i--){
36 res *= 5;
37 res += ss[i]-‘0‘;
38 }
39 return res;
40 }
41
42 void cal(string str,int r,int c,int wide){
43 if(is_black(r,c,wide)){
44 ans.push_back(consumption(str));
45 return;
46 }
47 else if(is_white(r,c,wide)) return;
48 else{
49 cal(str+"1",r,c,wide/2);
50 cal(str+"2",r,c+wide/2,wide/2);
51 cal(str+"3",r+wide/2,c,wide/2);
52 cal(str+"4",r+wide/2,c+wide/2,wide/2);
53 }
54 }
55
56 int solve(int n){
57 cnt = 0;
58 ans.clear();
59 for(int i = 0;i < n;i++){
60 scanf("%s",gra[i]);
61 }
62 string str = "";
63 if(is_black(0,0,n)) return 1;
64 if(is_white(0,0,n)) return -1;
65 cal(str+"1",0,0,n/2);
66 cal(str+"2",0,n/2,n/2);
67 cal(str+"3",n/2,0,n/2);
68 cal(str+"4",n/2,n/2,n/2);
69 return 0;
70 }
71
72 vector<int> num;
73
74 void converse(int val,string &ss){
75 while(val){
76 ss += (val%5+‘0‘);
77 val /= 5;
78 }
79 }
80
81 void draw(string &ss,int pos,int r,int c,int wide){
82 if(pos == ss.size()){
83 for(int i = r;i < r+wide;i++){
84 for(int j = c;j < c+wide;j++){
85 gra[i][j] = ‘*‘;
86 }
87 }
88 return;
89 }
90 if(ss[pos] == ‘1‘){
91 draw(ss,pos+1,r,c,wide/2);
92 }
93 else if(ss[pos] == ‘2‘){
94 draw(ss,pos+1,r,c+wide/2,wide/2);
95 }
96 else if(ss[pos] == ‘3‘){
97 draw(ss,pos+1,r+wide/2,c,wide/2);
98 }
99 else draw(ss,pos+1,r+wide/2,c+wide/2,wide/2);
100 }
101
102 void solve2(int n){
103 int x;
104 num.clear();
105 memset(gra,0,sizeof(gra));
106 while(scanf("%d",&x) && x!=-1) num.push_back(x);
107 if(num.size()==1 && num[0]==0){
108 for(int i = 0;i < n;i++){
109 for(int j = 0;j < n;j++){
110 printf("*");
111 }
112 printf("\n");
113 }
114 return;
115 }
116 for(int i = 0;i < num.size();i++){
117 string ss = "";
118 converse(num[i],ss);
119 draw(ss,0,0,0,n);
120 }
121 for(int i = 0;i < n;i++){
122 for(int j = 0;j < n;j++){
123 if(gra[i][j] == ‘*‘) printf("%c",gra[i][j]);
124 else printf(".");
125 }
126 printf("\n");
127 }
128 }
129
130 int iCase = 1;
131
132 int main()
133 {
134 //freopen("input.txt","r",stdin);
135 //freopen("output.txt","w",stdout);
136 bool flag = false;
137 while(~scanf("%d",&n) && n){
138 if(flag) printf("\n");
139 flag = true;
140 if(n > 0){
141 int flag = solve(n);
142 printf("Image %d\n",iCase++);
143 if(flag == 1){
144 printf("%d\n",0);
145 printf("Total number of black nodes = %d\n",1);
146 }
147 else if(flag == -1){
148 printf("Total number of black nodes = %d\n",0);
149 }
150 else{
151 sort(ans.begin(),ans.end());
152 int cnt = 0;
153 for(int i = 0;i < ans.size();i++){
154 if(cnt == 0) printf("%d",ans[i]);
155 else printf(" %d",ans[i]);
156 cnt++;
157 if(cnt == 12) cnt = 0,printf("\n");
158 }
159 if(ans.size()%12)printf("\n");
160 printf("Total number of black nodes = %d\n",ans.size());
161 }
162 }
163 else{
164 printf("Image %d\n",iCase++);
165 solve2(-n);
166 }
167 }
168 return 0;
169 }
原文地址：https://www.cnblogs.com/npugen/p/9527453.html