सी ++
कुल लंबाई: 9059, कुल क्षेत्रफल: 27469, विफलताएँ: 13।
नोट: स्कोर में विफलता दंड शामिल हैं।
नमूना उत्पादन:
a 6 b 8 c 11 d 11 e 11 f 11 g 12 h 15 i 18 j 18 k 21 l 23 m 26 n 28 o 28 p 30 q 32 r 33 s 33 t 34
------ae
| |
|---c
b||-g
|d|
f |
i---|
k---| h
| j
|---m
l | t
o-n|
|s-r
|-|
p q
Length: 39, Area: 150.
a 6 b 6 c 6 d 6 e 6 f 6 g 6 h 8 i 9 j 9 k 9 l 12 m 12 n 13 o 14 p 15 q 15 r 15 s 17 t 17 u 17 v 17 w 17 x 20 y 23 z 26
------a n|--w
|d-||---k|-o|
| g|b |--m --x
|-|c ||--r|
||f l|-q |
||--j u--|--s|-
e|-i |p| y|
h v t z-
Length: 56, Area: 120.
पूर्ण आउटपुट: http://pastebin.com/raw.php?i=spBUidBV
क्या आप सिर्फ जानवर बल समाधान प्यार नहीं करते? यह एक साधारण बैकग्राउंडिंग एल्गोरिथ्म से थोड़ा अधिक है: हमारे अथक कार्यकर्ता किसी भी बिंदु पर सभी संभावित चालों का परीक्षण करते समय, आवश्यक रूप से फ़्यूज़ और आतिशबाज़ी करते हुए, नक्शे के चारों ओर चलते हैं। खैर, लगभग --- हम चालों के सेट को प्रतिबंधित करते हैं और गैर-अपनाने वाले राज्यों को जल्दी छोड़ देते हैं ताकि यह असहनीय रूप से लंबा न हो (और, विशेष रूप से, ताकि यह समाप्त हो जाए।) किसी भी चक्र या अनपेक्षित को बनाने के लिए विशेष ध्यान नहीं दिया जाता है। रास्ते और हम जिस तरह से आए थे, उसी तरह वापस नहीं जाने के लिए, इसलिए यह गारंटी है कि हम एक ही राज्य में दो बार नहीं जाते हैं। फिर भी, एक इष्टतम समाधान खोजने में थोड़ा समय लग सकता है, इसलिए हम अंततः एक समाधान के अनुकूलन पर छोड़ देते हैं यदि यह बहुत लंबा लगता है।
इस एल्गोरिथ्म में अभी भी कुछ हेडरूम हैं। एक बात के लिए, FRUSTRATIONमापदंडों को बढ़ाकर बेहतर समाधान पाया जा सकता है । कोई प्रतिस्पर्धा एटीएम नहीं है, लेकिन इन नंबरों को क्रैंक किया जा सकता है अगर और जब ...
साथ संकलित करें: g++ fireworks.cpp -ofireworks -std=c++11 -pthread -O3।
साथ चलाएँ: ./fireworks।
एसटीडीआईएन से इनपुट पढ़ता है और एसटीडीयूएसटी (संभवतः आउट-ऑफ-ऑर्डर) के लिए आउटपुट लिखता है।
/* Magic numbers */
#define THREAD_COUNT 2
/* When FRUSTRATION_MOVES moves have passed since the last solution was found,
* the last (1-FRUSTRATION_STATES_BACKOFF)*100% of the backtracking states are
* discarded and FRUSTRATION_MOVES is multiplied by FRUSTRATION_MOVES_BACKOFF.
* The lower these values are, the faster the algorithm is going to give up on
* searching for better solutions. */
#define FRUSTRATION_MOVES 1000000
#define FRUSTRATION_MOVES_BACKOFF 0.8
#define FRUSTRATION_STATES_BACKOFF 0.5
#include <iostream>
#include <vector>
#include <algorithm>
#include <utility>
#include <thread>
#include <mutex>
#include <string>
#include <sstream>
#include <cassert>
using namespace std;
/* A tile on the board. Either a fuse, a firework, an empty tile or an
* out-of-boudns tile. */
struct tile {
/* The tile's value, encoded the "obvious" way (i.e. '-', '|', 'a', etc.)
* Empty tiles are encoded as '\0' and OOB tiles as '*'. */
char value;
/* For fuse tiles, the time at which the fuse is lit. */
int time;
operator char&() { return value; }
operator const char&() const { return value; }
bool is_fuse() const { return value == '-' || value == '|'; }
/* A tile is vacant if it's empty or OOB. */
bool is_vacant() const { return !value || value == '*'; }
/* Prints the tile. */
template <typename C, typename T>
friend basic_ostream<C, T>& operator<<(basic_ostream<C, T>& os,
const tile& t) {
return os << (t.value ? t.value : ' ');
}
};
/* Fireworks have the same encoding as tiles. */
typedef tile firework;
typedef vector<firework> fireworks;
/* The fuse map. It has physical dimensions (its bounding-box) but is
* conceptually infinite (filled with empty tiles.) */
class board {
/* The tiles, ordered left-to-right top-to-bottom. */
vector<tile> p_data;
/* The board dimensions. */
int p_width, p_height;
/* The total fuse length. */
int p_length;
public:
board(): p_width(0), p_height(0), p_length(0) {}
/* Physical dimensions. */
int width() const { return p_width; }
int height() const { return p_height; }
int area() const { return width() * height(); }
/* Total fuse length. */
int length() const { return p_length; }
/* Returns the tile at (x, y). If x or y are negative, returns an OOB
* tile. */
tile get(int x, int y) const {
if (x < 0 || y < 0)
return {'*'};
else if (x >= width() || y >= height())
return {'\0'};
else
return p_data[y * width() + x];
}
/* Sets the tile at (x, y). x and y must be nonnegative and the tile at
* (x, y) must be empty. */
board& set(int x, int y, const tile& t) & {
assert(x >= 0 && y >= 0);
assert(!get(x, y));
if (x >= width() || y >= height()) {
int new_width = x >= width() ? x + 1 : width();
int new_height = y >= height() ? y + 1 : height();
vector<tile> temp(new_width * new_height, {'\0'});
for (int l = 0; l < height(); ++l)
copy(
p_data.begin() + l * width(),
p_data.begin() + (l + 1) * width(),
temp.begin() + l * new_width
);
p_data.swap(temp);
p_width = new_width;
p_height = new_height;
}
p_data[y * width() + x] = t;
if (t.is_fuse())
++p_length;
return *this;
}
board&& set(int x, int y, const tile& t) && { return move(set(x, y, t)); }
/* Prints the board. */
template <typename C, typename T>
friend basic_ostream<C, T>& operator<<(basic_ostream<C, T>& os,
const board& b) {
for (int y = 0; y < b.height(); ++y) {
for (int x = 0; x < b.width(); ++x)
os << b.get(x, y);
os << endl;
}
return os;
}
};
/* A state of the tiling algorithm. */
struct state {
/* The current board. */
board b;
/* The next firework to tile. */
fireworks::const_iterator fw;
/* The current location. */
int x, y;
/* The current movement direction. 'N'orth 'S'outh 'E'ast, 'W'est or
* 'A'ny. */
char dir;
};
/* Adds a state to the state-stack if its total fuse length and bounding-box
* area are not worse than the current best ones. */
void add_state(vector<state>& states, int max_length, int max_area,
state&& new_s) {
if (new_s.b.length() < max_length ||
(new_s.b.length() == max_length && new_s.b.area() <= max_area)
)
states.push_back(move(new_s));
}
/* Adds the state after moving in a given direction, if it's a valid move. */
void add_movement(vector<state>& states, int max_length, int max_area,
const state& s, char dir) {
int x = s.x, y = s.y;
char parallel_fuse;
switch (dir) {
case 'E': if (s.dir == 'W') return; ++x; parallel_fuse = '|'; break;
case 'W': if (s.dir == 'E') return; --x; parallel_fuse = '|'; break;
case 'S': if (s.dir == 'N') return; ++y; parallel_fuse = '-'; break;
case 'N': if (s.dir == 'S') return; --y; parallel_fuse = '-'; break;
}
const tile t = s.b.get(s.x, s.y), nt = s.b.get(x, y);
assert(t.is_fuse());
if (nt.is_fuse() && !(t == parallel_fuse && nt == parallel_fuse))
add_state(states, max_length, max_area, {s.b, s.fw, x, y, dir});
}
/* Adds the state after moving in a given direction and tiling a fuse, if it's a
* valid move. */
void add_fuse(vector<state>& states, int max_length, int max_area,
const state& s, char dir, char fuse) {
int x = s.x, y = s.y;
int sgn;
bool horz;
switch (dir) {
case 'E': ++x; sgn = 1; horz = true; break;
case 'W': --x; sgn = -1; horz = true; break;
case 'S': ++y; sgn = 1; horz = false; break;
case 'N': --y; sgn = -1; horz = false; break;
}
if (s.b.get(x, y))
/* Tile is not empty. */
return;
/* Make sure we don't create cycles or reconnect a firework. */
const tile t = s.b.get(s.x, s.y);
assert(t.is_fuse());
if (t == '-') {
if (horz) {
if (fuse == '-') {
if (!s.b.get(x + sgn, y).is_vacant() ||
s.b.get(x, y - 1) == '|' ||
s.b.get(x, y + 1) == '|')
return;
} else {
if (s.b.get(x + sgn, y) == '-' ||
!s.b.get(x, y - 1).is_vacant() ||
!s.b.get(x, y + 1).is_vacant())
return;
}
} else {
if (!s.b.get(x, y + sgn).is_vacant() ||
s.b.get(x - 1, y) == '-' ||
s.b.get(x + 1, y) == '-')
return;
}
} else {
if (!horz) {
if (fuse == '|') {
if (!s.b.get(x, y + sgn).is_vacant() ||
s.b.get(x - 1, y) == '-' ||
s.b.get(x + 1, y) == '-')
return;
} else {
if (s.b.get(x, y + sgn) == '|' ||
!s.b.get(x - 1, y).is_vacant() ||
!s.b.get(x + 1, y).is_vacant())
return;
}
} else {
if (!s.b.get(x + sgn, y).is_vacant() ||
s.b.get(x, y - 1) == '|' ||
s.b.get(x, y + 1) == '|')
return;
}
}
/* Ok. */
add_state(
states,
max_length,
max_area,
{board(s.b).set(x, y, {fuse, t.time + 1}), s.fw, x, y, dir}
);
}
/* Adds the state after adding a firework at the given direction, if it's a
* valid move. */
void add_firework(vector<state>& states, int max_length, int max_area,
const state& s, char dir) {
int x = s.x, y = s.y;
int sgn;
bool horz;
switch (dir) {
case 'E': ++x; sgn = 1; horz = true; break;
case 'W': --x; sgn = -1; horz = true; break;
case 'S': ++y; sgn = 1; horz = false; break;
case 'N': --y; sgn = -1; horz = false; break;
}
if (s.b.get(x, y))
/* Tile is not empty. */
return;
/* Make sure we don't run into an undeliberate fuse. */
if (horz) {
if (s.b.get(x + sgn, y) == '-' || s.b.get(x, y - 1) == '|' ||
s.b.get(x, y + 1) == '|')
return;
} else {
if (s.b.get(x, y + sgn) == '|' || s.b.get(x - 1, y) == '-' ||
s.b.get(x + 1, y) == '-')
return;
}
/* Ok. */
add_state(
states,
max_length,
max_area,
/* After adding a firework, we can move in any direction. */
{board(s.b).set(x, y, {*s.fw}), s.fw + 1, s.x, s.y, 'A'}
);
}
void add_possible_moves(vector<state>& states, int max_length, int max_area,
const state& s) {
/* We add the new states in reverse-desirability order. The most
* (aesthetically) desirable states are added last. */
const tile t = s.b.get(s.x, s.y);
assert(t.is_fuse());
/* Move in all (possible) directions. */
for (char dir : "WENS")
if (dir) add_movement(states, max_length, max_area, s, dir);
/* If the fuse is too short for the next firework, keep adding fuse. */
if (t.time < s.fw->time) {
if (t == '-') {
add_fuse(states, max_length, max_area, s, 'N', '|');
add_fuse(states, max_length, max_area, s, 'S', '|');
add_fuse(states, max_length, max_area, s, 'W', '|');
add_fuse(states, max_length, max_area, s, 'W', '-');
add_fuse(states, max_length, max_area, s, 'E', '|');
add_fuse(states, max_length, max_area, s, 'E', '-');
} else {
add_fuse(states, max_length, max_area, s, 'W', '-');
add_fuse(states, max_length, max_area, s, 'E', '-');
add_fuse(states, max_length, max_area, s, 'N', '-');
add_fuse(states, max_length, max_area, s, 'N', '|');
add_fuse(states, max_length, max_area, s, 'S', '-');
add_fuse(states, max_length, max_area, s, 'S', '|');
}
} else if (t.time == s.fw->time) {
/* If we have enough fuse for the next firework, place the firework (if
* possible) and don't add more fuse, or else we'll never finish... */
if (t == '-') {
add_firework(states, max_length, max_area, s, 'W');
add_firework(states, max_length, max_area, s, 'E');
} else {
add_firework(states, max_length, max_area, s, 'N');
add_firework(states, max_length, max_area, s, 'S');
}
}
}
void thread_proc(mutex& lock, int& total_length, int& total_area,
int& failures) {
fireworks fw;
vector<state> states;
while (true) {
/* Read input. */
string input;
{
lock_guard<mutex> lg(lock);
while (!cin.eof() && input.empty())
getline(cin, input);
if (input.empty())
break;
}
fw.clear();
int length = 0, area;
{
stringstream is;
is << input;
while (!is.eof()) {
char c;
int t;
if (is >> c >> t) {
/* Fireworks must be sorted by launch time. */
assert(fw.empty() || t >= fw.back().time);
fw.push_back({c, t});
length += t;
}
}
assert(!fw.empty());
area = fw.back().time * fw.back().time;
}
/* Add initial state. */
states.push_back({board().set(0, 0, {'-', 1}), fw.begin(), 0, 0, 'A'});
board solution;
int moves = 0;
int frustration_moves = FRUSTRATION_MOVES;
while (!states.empty()) {
/* Check for solutions (all fireworks consumed.) */
while (!states.empty() && states.back().fw == fw.end()) {
state& s = states.back();
/* Did we find a better solution? */
if (solution.area() == 0 || s.b.length() < length ||
(s.b.length() == length && s.b.area() < area)
) {
solution = move(s.b);
moves = 0;
length = solution.length();
area = solution.area();
}
states.pop_back();
}
/* Expand the top state. */
if (!states.empty()) {
state s = move(states.back());
states.pop_back();
add_possible_moves(states, length, area, s);
}
/* Getting frustrated? */
++moves;
if (moves > frustration_moves) {
/* Get rid of some data. */
states.erase(
states.begin() + states.size() * FRUSTRATION_STATES_BACKOFF,
states.end()
);
frustration_moves *= FRUSTRATION_MOVES_BACKOFF;
moves = 0;
}
}
/* Print solution. */
{
lock_guard<mutex> lg(lock);
cout << input << endl;
if (solution.area())
cout << solution;
else {
cout << "FAILED!" << endl;
++failures;
}
cout << "Length: " << length <<
", Area: " << area <<
"." << endl << endl;
total_length += length;
total_area += area;
}
}
}
int main(int argc, const char* argv[]) {
thread threads[THREAD_COUNT];
mutex lock;
int total_length = 0, total_area = 0, failures = 0;
for (int i = 0; i < THREAD_COUNT; ++i)
threads[i] = thread(thread_proc, ref(lock), ref(total_length),
ref(total_area), ref(failures));
for (int i = 0; i < THREAD_COUNT; ++i)
threads[i].join();
cout << "Total Length: " << total_length <<
", Total Area: " << total_area <<
", Failures: " << failures <<
"." << endl;
}
अजगर
कुल लंबाई: 17387, कुल क्षेत्रफल: 62285, विफलताएँ: 44।
नमूना उत्पादन:
a 6 b 8 c 11 d 11 e 11 f 11 g 12 h 15 i 18 j 18 k 21 l 23 m 26 n 28 o 28 p 30 q 32 r 33 s 33 t 34
------a
|----f
|---c
b|||---h
|dg |
e |-j
|---k
i |
|---m
l |-o
|--p
n |--s
|-r
q|
t
Length: 45, Area: 345.
पूर्ण आउटपुट: http://pastebin.com/raw.php?i=mgiqXCRK
संदर्भ के लिए, यहाँ एक बहुत सरल तरीका है। यह आतिशबाजी को एक एकल मुख्य फ्यूज लाइन से जोड़ने की कोशिश करता है, जिससे "सीढ़ी" आकार बनता है। यदि एक फ़ायरवर्क मुख्य लाइन से सीधे कनेक्ट नहीं हो सकता (जो तब होता है जब एक ही समय में दो या दो से अधिक आतिशबाजी प्रकाश होता है) तो यह मुख्य लाइन को एक बिंदु की तलाश में वापस लाती है जहां यह लंबवत रूप से नीचे या दाईं ओर शाखा कर सकती है (और विफल होने पर) ऐसी कोई बात मौजूद नहीं है।)
अप्रत्याशित रूप से, यह ब्रूट-फोर्स सॉल्वर से भी बदतर है, लेकिन एक बड़े अंतर से नहीं। ईमानदारी से, मुझे उम्मीद है कि अंतर कुछ बड़ा होगा।
साथ चलाएँ: python fireworks.py।
from __future__ import print_function
import sys
total_length = total_area = failures = 0
for line in sys.stdin:
# Read input.
line = line.strip()
if line == "": continue
fws = line.split(' ')
# The fireworks are a list of pairs of the form (<letter>, <time>).
fws = [(fws[i], int(fws[i + 1])) for i in xrange(0, len(fws), 2)]
# The board is a dictionary of the form <coord>: <tile>.
# The first tile marks the "starting point" and is out-of-bounds.
board = {(-1, 0): '*'}
# The tip of the main "staircase" fuse.
tip_x, tip_y = -1, 0
tip_time = 0
# We didn't fail. Yet...
failed = False
for (fw, fw_time) in fws:
dt = fw_time - tip_time
# Can we add the firework to the main fuse line?
if dt > 0:
# We can. Alternate the direction to create a "staircase" pattern.
if board[(tip_x, tip_y)] == '-': dx, dy = 0, 1; fuse = '|'
else: dx, dy = 1, 0; fuse = '-'
x, y = tip_x, tip_y
tip_x += dt * dx
tip_y += dt * dy
tip_time += dt
else:
# We can't. Trace the main fuse back until we find a point where we
# can thread, or fail if we reach the starting point.
x, y = tip_x, tip_y
while board[(x, y)] != '*':
horz = board[(x, y)] == '-'
if horz: dx, dy = 0, 1; fuse = '|'
else: dx, dy = 1, 0; fuse = '-'
if dt > 0 and (x + dx, y + dy) not in board: break
if horz: x -= 1
else: y -= 1
dt += 1
if board[(x, y)] == '*':
failed = True
break
# Add the fuse and firework.
for i in xrange(dt):
x += dx; y += dy
board[(x, y)] = fuse
board[(x + dx, y + dy)] = fw
# Print output.
print(line)
if not failed:
max_x, max_y = (max(board, key=lambda p: p[i])[i] + 1 for i in (0, 1))
for y in xrange(max_y):
for x in xrange(max_x):
print(board.get((x, y), ' '), end = "")
print()
length = len(board) - len(fws) - 1
area = max_x * max_y
else:
print("FAILED!")
failures += 1
length = sum(map(lambda fw: fw[1], fws))
area = fws[-1][1] ** 2
print("Length: %d, Area: %d.\n" % (length, area))
total_length += length; total_area += area
print("Total Length: %d, Total Area: %d, Failures: %d." %
(total_length, total_area, failures))