USACO: Packing Rectangles

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#basic #output #each #layout #integer #parallel

本文解析了一道经典的矩形排列模拟题，旨在寻找四个矩形以最小面积组合的方式。通过对不同排列组合的分析，提供了完整的C++实现代码，并通过实际测试案例验证了程序的有效性和正确性。

Packing Rectangles
IOI 95

The six basic layouts of four rectangles

Four rectangles are given. Find the smallest enclosing (new) rectangle into which these four may be fitted without overlapping. By smallest rectangle, we mean the one with the smallest area.

All four rectangles should have their sides parallel to the corresponding sides of the enclosing rectangle. Figure 1 shows six ways to fit four rectangles together. These six are the only possible basic layouts, since any other layout can be obtained from a basic layout by rotation or reflection. Rectangles may be rotated 90 degrees during packing.

There may exist several different enclosing rectangles fulfilling the requirements, all with the same area. You must produce all such enclosing rectangles.

PROGRAM NAME: packrec

INPUT FORMAT

Four lines, each containing two positive space-separated integers that represent the lengths of a rectangle's two sides. Each side of a rectangle is at least 1 and at most 50.

SAMPLE INPUT (file packrec.in)

OUTPUT FORMAT

The output file contains one line more than the number of solutions. The first line contains a single integer: the minimum area of the enclosing rectangles. Each of the following lines contains one solution described by two numbers p and q with p<=q. These lines must be sorted in ascending order of p, and must all be different.

SAMPLE OUTPUT (file packrec.out)

40
4 10
5 8

完完全全的模拟题，属于那种做了就不想再做一次的题目……

/* ID: geterns1 PROG: packrec LANG: C++ */ #include <iostream> #include <fstream> using namespace std; #define size 4 struct rct{ int len, wth; int area(){ return len * wth; } }; rct base1(int odr1[], int odr2[], rct rcts[][size]){ rct result; result.len = 0; result.wth = 0; for (int i = 0; i < 4; i ++){ result.len += rcts[odr2[i]][odr1[i]].len; if (rcts[odr2[i]][odr1[i]].wth > result.wth) result.wth = rcts[odr2[i]][odr1[i]].wth; } return result; } rct base2(int odr1[], int odr2[], rct rcts[][size]){ rct result; result.len = 0; result.wth = 0; for (int i = 0; i < 3; i ++){ result.len += rcts[odr2[i]][odr1[i]].len; if (rcts[odr2[i]][odr1[i]].wth > result.wth) result.wth = rcts[odr2[i]][odr1[i]].wth; } if (rcts[odr2[3]][odr1[3]].len > result.len) result.len = rcts[odr2[3]][odr1[3]].len; result.wth += rcts[odr2[3]][odr1[3]].wth; return result; } rct base3(int odr1[], int odr2[], rct rcts[][size]){ rct result; result.len = rcts[odr2[0]][odr1[0]].len + rcts[odr2[1]][odr1[1]].len; if (rcts[odr2[2]][odr1[2]].len > result.len) result.len = rcts[odr2[2]][odr1[2]].len; result.len += rcts[odr2[3]][odr1[3]].len; if (rcts[odr2[0]][odr1[0]].wth > rcts[odr2[1]][odr1[1]].wth) result.wth = rcts[odr2[0]][odr1[0]].wth; else result.wth = rcts[odr2[1]][odr1[1]].wth; result.wth += rcts[odr2[2]][odr1[2]].wth; if (rcts[odr2[3]][odr1[3]].wth > result.wth) result.wth = rcts[odr2[3]][odr1[3]].wth; return result; } rct base4(int odr1[], int odr2[], rct rcts[][size]){ rct result; if (rcts[odr2[1]][odr1[1]].len > rcts[odr2[2]][odr1[2]].len) result.len = rcts[odr2[1]][odr1[1]].len; else result.len = rcts[odr2[2]][odr1[2]].len; result.len = result.len + rcts[odr2[0]][odr1[0]].len + rcts[odr2[3]][odr1[3]].len; result.wth = rcts[odr2[1]][odr1[1]].wth + rcts[odr2[2]][odr1[2]].wth; if (rcts[odr2[0]][odr1[0]].wth > result.wth) result.wth = rcts[odr2[0]][odr1[0]].wth; if (rcts[odr2[3]][odr1[3]].wth > result.wth) result.wth = rcts[odr2[3]][odr1[3]].wth; return result; } rct base5(int odr1[], int odr2[], rct rcts[][size]){ rct result; if (rcts[odr2[0]][odr1[0]].len > rcts[odr2[1]][odr1[1]].len) result.len = rcts[odr2[0]][odr1[0]].len; else result.len = rcts[odr2[1]][odr1[1]].len; result.len = result.len + rcts[odr2[2]][odr1[2]].len + rcts[odr2[3]][odr1[3]].len; result.wth = rcts[odr2[0]][odr1[0]].wth + rcts[odr2[1]][odr1[1]].wth; if (rcts[odr2[2]][odr1[2]].wth > result.wth) result.wth = rcts[odr2[2]][odr1[2]].wth; if (rcts[odr2[3]][odr1[3]].wth > result.wth) result.wth = rcts[odr2[3]][odr1[3]].wth; return result; } rct base6(int odr1[], int odr2[], rct rcts[][size]){ rct result; int len1 = rcts[odr2[0]][odr1[0]].len + rcts[odr2[3]][odr1[3]].len; int len2 = rcts[odr2[1]][odr1[1]].len + rcts[odr2[2]][odr1[2]].len; result.wth = rcts[odr2[0]][odr1[0]].wth + rcts[odr2[1]][odr1[1]].wth; if (rcts[odr2[2]][odr1[2]].wth + rcts[odr2[3]][odr1[3]].wth > result.wth) result.wth = rcts[odr2[2]][odr1[2]].wth + rcts[odr2[3]][odr1[3]].wth; if (len1 >= len2){ result.len = len1; if (rcts[odr2[1]][odr1[1]].len > rcts[odr2[0]][odr1[0]].len) if (rcts[odr2[1]][odr1[1]].wth + rcts[odr2[3]][odr1[3]].wth > result.wth) result.wth = rcts[odr2[1]][odr1[1]].wth + rcts[odr2[3]][odr1[3]].wth; if (rcts[odr2[2]][odr1[2]].len > rcts[odr2[3]][odr1[3]].len) if (rcts[odr2[0]][odr1[0]].wth + rcts[odr2[2]][odr1[2]].wth > result.wth) result.wth = rcts[odr2[0]][odr1[0]].wth + rcts[odr2[2]][odr1[2]].wth; } else{ result.len = len2; if (rcts[odr2[1]][odr1[1]].len < rcts[odr2[0]][odr1[0]].len) if (rcts[odr2[0]][odr1[0]].wth + rcts[odr2[2]][odr1[2]].wth > result.wth) result.wth = rcts[odr2[0]][odr1[0]].wth + rcts[odr2[2]][odr1[2]].wth; if (rcts[odr2[2]][odr1[2]].len < rcts[odr2[3]][odr1[3]].len) if (rcts[odr2[1]][odr1[1]].wth + rcts[odr2[3]][odr1[3]].wth > result.wth) result.wth = rcts[odr2[1]][odr1[1]].wth + rcts[odr2[3]][odr1[3]].wth; } return result; } int main(){ ifstream ifs("packrec.in"); ofstream ofs("packrec.out"); rct rcts[2][size]; for (int i = 0; i < size; i ++){ ifs >> rcts[0][i].len >> rcts[0][i].wth; rcts[1][i].len = rcts[0][i].wth; rcts[1][i].wth = rcts[0][i].len; } int order1[24][size] = { { 0, 1, 2, 3 }, { 0, 1, 3, 2 }, { 0, 2, 1, 3 }, { 0, 2, 3, 1 }, { 0, 3, 1, 2 }, { 0, 3, 2, 1 }, { 1, 0, 2, 3 }, { 1, 0, 3, 2 }, { 1, 2, 0, 3 }, { 1, 2, 3, 0 }, { 1, 3, 0, 2 }, { 1, 3, 2, 0 }, { 2, 0, 1, 3 }, { 2, 0, 3, 1 }, { 2, 1, 0, 3 }, { 2, 1, 3, 0 }, { 2, 3, 0, 1 }, { 2, 3, 1, 0 }, { 3, 0, 1, 2 }, { 3, 0, 2, 1 }, { 3, 1, 0, 2 }, { 3, 1, 2, 0 }, { 3, 2, 0, 1 }, { 3, 2, 1, 0 } }; int order2[16][size] = { { 0, 0, 0, 0 }, { 0, 0, 0, 1 }, { 0, 0, 1, 0 }, { 0, 0, 1, 1 }, { 0, 1, 0, 0 }, { 0, 1, 0, 1 }, { 0, 1, 1, 0 }, { 0, 1, 1, 1 }, { 1, 0, 0, 0 }, { 1, 0, 0, 1 }, { 1, 0, 1, 0 }, { 1, 0, 1, 1 }, { 1, 1, 0, 0 }, { 1, 1, 0, 1 }, { 1, 1, 1, 0 }, { 1, 1, 1, 1 } }; rct (* funPtr[6])( int[], int[], rct[][size]) = { base1, base2, base3, base4, base5, base6 }; rct tmpRct, minRcts[1000]; int tmpArea, minArea = 10001, counter; for (int i = 0; i < 24; i ++){ for (int j = 0; j < 16; j ++){ for (int k = 0; k < 6; k ++){ tmpRct = funPtr[k](order1[i], order2[j], rcts); tmpArea = tmpRct.area(); if (tmpArea < 40) int a = 0; if (tmpArea < minArea){ counter = 0; minRcts[counter] = tmpRct; minArea = tmpArea; continue; } if (tmpArea == minArea && tmpRct.len != minRcts[counter].len && tmpRct.len != minRcts[counter].wth) { counter ++; minRcts[counter] = tmpRct; } } } } //resort elements for (int i = 0; i <= counter; i ++){ if (minRcts[i].len > minRcts[i].wth){ int tmp = minRcts[i].len; minRcts[i].len = minRcts[i].wth; minRcts[i].wth = tmp; } } //sort by selecting for (int i = 0; i <= counter; i ++){ int min = 100, index; for (int j = i; j <= counter; j ++){ if (minRcts[j].len < min){ min = minRcts[j].len; index = j; } } if (index == i) continue; rct tmp; tmp = minRcts[i]; minRcts[i] = minRcts[index]; minRcts[index] = tmp; } ofs << minArea << endl; for (int i = 0; i <= counter; i ++){ if (i != 0 && minRcts[i].len == minRcts[i - 1].len) continue; ofs << minRcts[i].len << " " << minRcts[i].wth << endl; } ifs.close(); ofs.close(); return 0; }

USER: Geterns Liu [geterns1]
TASK: packrec
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.022 secs, 3016 KB]
   Test 2: TEST OK [0.000 secs, 3016 KB]
   Test 3: TEST OK [0.000 secs, 3016 KB]
   Test 4: TEST OK [0.011 secs, 3016 KB]
   Test 5: TEST OK [0.011 secs, 3016 KB]
   Test 6: TEST OK [0.000 secs, 3016 KB]
   Test 7: TEST OK [0.000 secs, 3016 KB]
   Test 8: TEST OK [0.000 secs, 3016 KB]
   Test 9: TEST OK [0.000 secs, 3016 KB]
   Test 10: TEST OK [0.011 secs, 3016 KB]
   Test 11: TEST OK [0.011 secs, 3016 KB]
   Test 12: TEST OK [0.011 secs, 3016 KB]
   Test 13: TEST OK [0.000 secs, 3016 KB]
   Test 14: TEST OK [0.000 secs, 3016 KB]
   Test 15: TEST OK [0.000 secs, 3016 KB]
   Test 16: TEST OK [0.011 secs, 3016 KB]
   Test 17: TEST OK [0.000 secs, 3016 KB]
   Test 18: TEST OK [0.000 secs, 3016 KB]
   Test 19: TEST OK [0.022 secs, 3016 KB]
   Test 20: TEST OK [0.000 secs, 3016 KB]
   Test 21: TEST OK [0.000 secs, 3016 KB]

All tests OK.
Your program ('packrec') produced all correct answers!  This is your
submission #8 for this problem.  Congratulations!

Here are the test data inputs:

------- test 1 -------
1 2
2 3
3 4
4 5
------- test 2 -------
20 20
20 20
20 20
20 20
------- test 3 -------
4 5
5 4
4 5
16 1
------- test 4 -------
4 5
2 5
5 2
2 10
------- test 5 -------
12 18
4 6
2 17
19 3
------- test 6 -------
10 10
5 5
15 15
20 20
------- test 7 -------
1 1
1 20
1 20
20 20
------- test 8 -------
5 8
3 12
15 4
14 10
------- test 9 -------
4 5
5 6
6 4
4 5
------- test 10 -------
1 5
5 10
10 15
15 20
------- test 11 -------
3 4
8 5
7 1
4 5
------- test 12 -------
17 11
16 20
4 6
13 19
------- test 13 -------
4 2
2 6
2 3
5 8
------- test 14 -------
1 2
2 3
3 4
4 5
------- test 15 -------
4 8
8 12
12 16
16 20
------- test 16 -------
3 5
1 3
2 4
2 5
------- test 17 -------
4 3
4 4
6 3
5 5
------- test 18 -------
49 50
49 50
49 50
49 50
------- test 19 -------
10 50
45 30
28 38
36 20
------- test 20 -------
50 49
49 48
48 47
47 46
------- test 21 -------
50 49
48 47
46 45
45 44

Keep up the good work!

Thanks for your submission!