UVa - 103 - Stacking Boxes
2015-07-02 20:04
489 查看
Background
Some concepts in Mathematics and Computer Science are simple in one or two dimensions but become more complex when extended to arbitrary dimensions. Consider solving differential equations in several dimensionsand analyzing the topology of an n-dimensional hypercube. The former is much more complicated than its one dimensional relative while the latter bears a remarkable resemblance to its ``lower-class'' cousin.
The Problem
Consider an n-dimensional ``box'' given by its dimensions. In two dimensions the box (2,3) might represent a box with length 2 units and width 3 units. In three dimensions the box (4,8,9) can representa box
(length, width, and height). In 6 dimensions it is, perhaps, unclear what the box (4,5,6,7,8,9) represents; but we can analyze
properties of the box such as the sum of its dimensions.
In this problem you will analyze a property of a group of n-dimensional boxes. You are to determine the longest nesting string of boxes, that is a sequence of boxes
such
that each box
nests in box
(
.
A box D = (
) nests in a box E = (
)
if there is some rearrangement of the
such that when rearranged each dimension is less than the corresponding dimension in box
E. This loosely corresponds to turning box D to see if it will fit in box E. However, since any rearrangement suffices, box D can be contorted, not just turned (see examples below).
For example, the box D = (2,6) nests in the box E = (7,3) since D can be rearranged as (6,2) so that each dimension is less than the corresponding dimension in E. The box D = (9,5,7,3) does NOT nest in the box E
= (2,10,6,8) since no rearrangement of D results in a box that satisfies the nesting property, but F = (9,5,7,1) does nest in box E since F can be rearranged as (1,9,5,7) which nests in E.
Formally, we define nesting as follows: box D = (
) nests in box E = (
)
if there is a permutation
of
such
that (
) ``fits'' in (
)
i.e., if
for all
.
The Input
The input consists of a series of box sequences. Each box sequence begins with a line consisting of the the number of boxes k in the sequence followed by the dimensionality of the boxes, n (onthe same line.)
This line is followed by k lines, one line per box with the n measurements of each box on one line separated by one or more spaces. The
line
in the sequence (
) gives the measurements for the
box.
There may be several box sequences in the input file. Your program should process all of them and determine, for each sequence, which of the k boxes determine the longest nesting string and the length of
that nesting string (the number of boxes in the string).
In this problem the maximum dimensionality is 10 and the minimum dimensionality is 1. The maximum number of boxes in a sequence is 30.
The Output
For each box sequence in the input file, output the length of the longest nesting string on one line followed on the next line by a list of the boxes that comprise this string in order. The ``smallest'' or ``innermost''box of the nesting string should be listed first, the next box (if there is one) should be listed second, etc.
The boxes should be numbered according to the order in which they appeared in the input file (first box is box 1, etc.).
If there is more than one longest nesting string then any one of them can be output.
Sample Input
5 2 3 7 8 10 5 2 9 11 21 18 8 6 5 2 20 1 30 10 23 15 7 9 11 3 40 50 34 24 14 4 9 10 11 12 13 14 31 4 18 8 27 17 44 32 13 19 41 19 1 2 3 4 5 6 80 37 47 18 21 9
Sample Output
5 3 1 2 4 5 4 7 2 5 6
动态规划。和嵌套矩形问题本质是一样的,不过这里不需要输出字典序最小的结果。
AC代码:
#include <iostream> #include <cstdio> #include <cstdlib> #include <cctype> #include <cstring> #include <string> #include <sstream> #include <vector> #include <set> #include <map> #include <algorithm> #include <stack> #include <queue> #include <bitset> #include <cassert> #include <cmath> #include <functional> using namespace std; const int maxk = 35; int k, n, ans, target; int dp[maxk]; bool first; struct Box { int dimension[12]; bool operator < (const Box& rhs) const { // 方便判断是否嵌套 for (int i = 0; i < n; i++) { if (dimension[i] >= rhs.dimension[i]) { return false; } } return true; } }box[maxk]; void init() { memset(dp, -1, sizeof(dp)); first = true; ans = -1; cin >> n; for (int i = 0; i < k; i++) { for (int j = 0; j < n; j++) { cin >> box[i].dimension[j]; } sort(box[i].dimension, box[i].dimension + n); } } int dfs(int i, const vector<int> *mp) { int &ans = dp[i], temp; if (ans != -1) { return ans; } ans = 1; for (int j = 0; j < mp[i].size(); j++) { temp = dfs(mp[i][j], mp) + 1; ans = max(temp, ans); } return ans; } void printAns(int i, const vector<int> *mp) { if (first) { cout << i + 1; first = false; } else { cout << ' ' << i + 1; } for (int j = 0; j < mp[i].size(); j++) { if (dp[mp[i][j]] == dp[i] - 1) { printAns(mp[i][j], mp); break; } } } void solve() { vector<int> mp[maxk]; for (int i = 0; i < k - 1; i++) { for (int j = i + 1; j < k; j++) { if (box[i] < box[j]) { mp[i].push_back(j); } else if (box[j] < box[i]) { mp[j].push_back(i); } } } for (int i = 0; i < k; i++) { int temp = dfs(i, mp); if (temp > ans) { ans = temp; target = i; } } cout << ans << endl; printAns(target, mp); cout << endl; } int main() { ios::sync_with_stdio(false); while (cin >> k) { init(); solve(); } return 0; }
相关文章推荐
- java 基础知识随笔
- mysql登录报错 ERROR 1045 (28000)
- codeforces #311 557D D. Vitaly and Cycle (dfs+图论判断判断构造奇环)
- NYOJ 103 a+b Problem(2)
- 桥接模式
- Git学习 - 【git】SourceTreeで解説!間違えたときのgitコマンドとやり方
- 2014成都百万职工职业技能大赛计算机程序员决赛在四川华迪顺利举行
- rman恢复增加ADG(级联ADG)
- 《Java程序实习》日记(周四)
- 安卓作业—显示图片
- 黑马程序员——自增和自减运算符
- 如何使用Android Studio把自己的Android library分发到jCenter和Maven Central
- 35.Search Insert Position
- #Shell脚本--输出两个字符串的最长匹配部分
- MyBatis数据持久化(十一)Mybatis3、Spring4、Struts2整合开发
- MyBatis数据持久化(十一)Mybatis3、Spring4、Struts2整合开发
- Search in Rotated Sorted Array
- MyBatis数据持久化(十一)Mybatis3、Spring4、Struts2整合开发
- 挑逗B少年搞计划10 假设你是愿意用我的心脏层剥离一层~
- 2015比较常用的第三方类库