codeforces #322 div 2 D. Three Logos (枚举)
D. Three Logos
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Three companies decided to order a billboard with pictures of their logos. A billboard is a big square board. A logo of each company is a rectangle of a non-zero area.
Advertisers will put up the ad only if it is possible to place all three logos on the billboard so that they do not overlap and the billboard has no empty space left. When you put a logo on the billboard, you should rotate it so that the sides were parallel to the sides of the billboard.
Your task is to determine if it is possible to put the logos of all the three companies on some square billboard without breaking any of the described rules.
Input
The first line of the input contains six positive integers _x_1, _y_1, _x_2, _y_2, _x_3, _y_3 (1 ≤ _x_1, _y_1, _x_2, _y_2, _x_3, _y_3 ≤ 100), where x__i and y__i determine the length and width of the logo of the i-th company respectively.
Output
If it is impossible to place all the three logos on a square shield, print a single integer “-1” (without the quotes).
If it is possible, print in the first line the length of a side of square n, where you can place all the three logos. Each of the next n lines should contain n uppercase English letters “A”, “B” or “C”. The sets of the same letters should form solid rectangles, provided that:
- the sizes of the rectangle composed from letters “A” should be equal to the sizes of the logo of the first company,
- the sizes of the rectangle composed from letters “B” should be equal to the sizes of the logo of the second company,
- the sizes of the rectangle composed from letters “C” should be equal to the sizes of the logo of the third company,
Note that the logos of the companies can be rotated for printing on the billboard. The billboard mustn’t have any empty space. If a square billboard can be filled with the logos in multiple ways, you are allowed to print any of them.
See the samples to better understand the statement.
Sample test(s)
input
5 1 2 5 5 2
output
5<br></br>AAAAA<br></br>BBBBB<br></br>BBBBB<br></br>CCCCC<br></br>CCCCC
input
4 4 2 6 4 2
output
6<br></br>BBBBBB<br></br>BBBBBB<br></br>AAAACC<br></br>AAAACC<br></br>AAAACC<br></br>AAAACC<br></br><br></br>题意很清楚．<br></br>给三个矩形，问能否拼成一个正方形．<br></br>我们先把矩形都躺着放（保证x<=y）<br></br>然后容易知道，一共有两种可能的方法<br></br>一种是三个矩形罗在一起<br></br>另外一种是一个矩形在上面，下面两个矩形竖着放．<br></br>后一种方法又因为上面的矩形的不同以及下面两个矩形的横竖放置不同（00 01 10 11）＊３<br></br>所以一共有１２种要枚举...<br></br>直接写会累死...<br></br>想了个简单点的办法，具体见代码．<br></br>一遍AC,好爽！！！
1/*************************************************************************
2 > File Name: code/cf/#322/D.cpp
3 > Author: 111qqz
4 > Email: rkz2013@126.com
5 > Created Time: 2015年09月30日 星期三 17时21分24秒
6 ************************************************************************/
7#include<iostream>
8#include<iomanip>
9#include<cstdio>
10#include<algorithm>
11#include<cmath>
12#include<cstring>
13#include<string>
14#include<map>
15#include<set>
16#include<queue>
17#include<vector>
18#include<stack>
19#include<cctype>
20#define y1 hust111qqz
21#define yn hez111qqz
22#define j1 cute111qqz
23#define ms(a,x) memset(a,x,sizeof(a))
24#define lr dying111qqz
25using namespace std;
26#define For(i, n) for (int i=0;i<int(n);++i)
27typedef long long LL;
28typedef double DB;
29const int inf = 0x3f3f3f3f;
30int ans;
31struct Point
32{
33 int x,y;
34}p[10];
35char maze[110][110];
36int ss;
37int lst;
38int a[10],b[10];
39bool solve()
40{
41 if (lst==-1) return false;
42 // cout<<"lst:"<<lst<<endl;
43 if (p[0].x+p[1].x+p[2].x==p[0].y&&p[0].y==p[1].y&&p[0].y==p[2].y)
44 {
45 ans = p[0].y;
46 for ( int i = 0 ; i < p[0].x ; i++)
47 {
48 for ( int j = 0 ; j < ans ; j++)
49 {
50 maze[i][j] = 'A';
51 }
52 }
53 for ( int i = p[0].x ; i < p[0].x+p[1].x;i++)
54 {
55 for ( int j = 0 ; j < ans ; j++)
56 {
57 maze[i][j] = 'B';
58 }
59 }
60 for ( int i = p[0].x + p[1].x ; i < ans ; i++)
61 {
62 for ( int j = 0 ; j < ans ; j++)
63 {
64 maze[i][j] = 'C';
65 }
66 }
67 return true;
68 }
69
70 if (p[a[lst]].x+p[b[lst]].x==p[lst].y&&p[a[lst]].y==p[b[lst]].y&&p[a[lst]].y+p[lst].x==p[lst].y)
71 {
72 ans = p[lst].y;
73 for ( int i = 0 ; i < p[lst].x ; i++)
74 {
75 for ( int j = 0 ; j < ans ; j++)
76 {
77 maze[i][j] = char(lst+'A');
78 }
79 }
80 for ( int i = p[lst].x ; i < ans ; i++)
81 {
82 for ( int j = 0 ; j < ans ;j++)
83 {
84 if (j<p[a[lst]].x)
85 {
86 maze[i][j]=char(a[lst]+'A');
87 // cout<<i<<" "<<j<<" "<<char(a[lst]+'A')<<endl;
88 }
89 else
90 {
91 maze[i][j] = char (b[lst] + 'A');
92 }
93 }
94 }
95 return true;
96 }
97
98
99 if (p[a[lst]].x+ p[b[lst]].y==p[lst].y&&p[a[lst]].y==p[b[lst]].x&&p[a[lst]].y+p[lst].x==p[lst].y)
100 {
101 //cout<<"22222222222222222"<<endl;
102 ans = p[lst].y;
103 for ( int i = 0 ; i < p[lst].x ; i++)
104 {
105 for ( int j = 0 ; j < ans ; j ++)
106 {
107 maze[i][j] = char(lst+'A');
108 }
109 }
110
111 for ( int i = p[lst].x ; i < ans ; i++)
112 {
113 for ( int j = 0 ; j < ans ; j++)
114 {
115 if (j<p[a[lst]].x)
116 {
117 maze[i][j] =char(a[lst]+'A');
118 }
119 else
120 {
121 maze[i][j] = char(b[lst]+'A');
122 }
123 }
124 }
125 return true;
126 }
127if (p[a[lst]].y+p[b[lst]].x==p[lst].y&&p[a[lst]].x==p[b[lst]].y&&p[a[lst]].x+p[lst].x==p[lst].y)
128 {
129 //cout<<"3333333333333333333333"<<endl;
130 ans = p[lst].y;
131 for ( int i = 0 ; i < p[lst].x ; i++)
132 {
133 for ( int j = 0 ; j < ans ; j++)
134 {
135 maze[i][j] = char(lst+'A');
136 }
137 }
138 for ( int i = p[lst].x ; i < ans ; i++)
139 {
140 for ( int j = 0 ; j < ans ;j++)
141 {
142 if (j<p[a[lst]].y)
143 {
144 maze[i][j]=char(a[lst]+'A');
145 }
146 else
147 {
148 maze[i][j] = char (b[lst] + 'A');
149 }
150 }
151 }
152 return true;
153 }
154if (p[a[lst]].y+p[b[lst]].y==p[lst].y&&p[a[lst]].x==p[b[lst]].x&&p[a[lst]].x+p[lst].x==p[lst].y)
155 {
156 // cout<<"44444444444444444444444444"<<endl;
157 ans = p[lst].y;
158 for ( int i = 0 ; i < p[lst].x ; i++)
159 {
160 for ( int j = 0 ; j < ans ; j++)
161 {
162 maze[i][j] = char(lst+'A');
163 }
164 }
165 for ( int i = p[lst].x ; i < ans ; i++)
166 {
167 for ( int j = 0 ; j < ans ;j++)
168 {
169 if (j<p[a[lst]].y)
170 {
171 maze[i][j]=char(a[lst]+'A');
172 // cout<<"haaa"<<a[lst]+'A'<<endl;
173 }
174 else
175 {
176 maze[i][j] = char (b[lst] + 'A');
177 }
178 }
179 }
180 return true;
181 }
182 return false;
183}
184int main()
185{
186 #ifndef ONLINE_JUDGE
187 freopen("in.txt","r",stdin);
188 #endif
189
190 ss = 0;
191 lst = -1;
192 a[0]=1;b[0]=2;
193 a[1]=0;b[1]=2;
194 a[2]=0;b[2]=1;
195 for ( int i = 0 ; i < 3 ; i++)
196 {
197 scanf("%d %d",&p[i].x,&p[i].y);
198 if (p[i].x>p[i].y) swap(p[i].x,p[i].y);
199 ss += p[i].x*p[i].y;
200 }
201 for ( int i = 0 ; i < 3 ; i++)
202 {
203 if (p[i].y*p[i].y==ss)
204 {
205 lst = i;
206 }
207 }
208 if (solve())
209 {
210 printf("%d\n",ans);
211 for ( int i = 0 ; i < ans ; i++)
212 {
213 for ( int j = 0 ; j < ans ; j++)
214 {
215 printf("%c",maze[i][j]);
216 }
217 printf("\n");
218 }
219 }
220 else
221 {
222 puts("-1");
223 }
224 #ifndef ONLINE_JUDGE
225
226 fclose(stdin);
227 #endif
228 return 0;
229}