## Friday, October 3, 2014

### UVa Problem 10360 - Rat Attack

Problem:

Solution:

I started to attempt this problem without reading the solution mentioned in competitive programming, but I read the hint that this is brute force and backward. How backward exactly?

First, the frontal attack is obviously bombing all position. A quick analysis shows it takes 1025 x 1025 bombing, and each bombing takes number of rat population time to check how many rats are killed. That is a huge number of memory access.

First observation is that there is a lot of bombing required - and most of them are futile since they are likely bombing area with no rats at all. So I thought, we can choose to bomb area with only rats. From the rats, we find the area we need to bomb. So that becomes 20000 x (2 x strength + 1) x (2 x strength + 1) bombs. That reduce the number of bombs. However, each bomb is still taking number of rat population time.

I tried this approach, still getting time limit exceeded.

So I tried a couple approaches to save more time. First, I avoid bombing the same position. If two rats are nearby, the loop might bomb exactly the same position. Second, I used kd-tree to speed up bombing speed to log(number of rat populations). That is an orthogonal range search problem.

The code is quite complex, and too bad, it causes run-time error on judge. Till now I don't know what's wrong with my code. I tried some fairly random input and is reporting the same result as the original search.

At this point, I peeked at the solution on the book. The solution on book is really cool because it is a backward direction I never thought of. Instead of asking how many rats a bomb can kill, we ask, which bomb positions can kill the rats. By storing how many rats are killed in each cell, a simple search can solve the problem.

The book solution got accepted, I didn't bother to fix the kd-tree more, but I bet we will need it in other problems, so I keep it. C macros are pretty useful here.

What I learnt:

• Try re-order things - answer might just pop up.
• kd-tree median logic can be quite complicated - we must make sure we split the x list and y list the same way - otherwise it will go into stack overflow.
• If I need to write it again, make sure I try test case with identical median x and median y values.

Code:

#include "stdafx.h"

// http://uva.onlinejudge.org/index.php?option=onlinejudge&page=show_problem&problem=1301

#include "UVa10360.h"

#include <iostream>
#include <list>
#include <vector>
#include <algorithm>
#include <cmath>
#include <map>

using namespace std;

struct rat_population
{
int x;
int y;
int size;
};

#define UVa10360_USEKDTREE

#ifdef UVa10360_USEKDTREE

struct kdtree_node
{
bool by_x;
bool is_leaf;
double split_value;
rat_population* data;
kdtree_node* left;
kdtree_node* right;
kdtree_node() : left(NULL), right(NULL), data(NULL), is_leaf(false)
{
}
~kdtree_node()
{
if (this->left != NULL)
{
delete this->left;
}
if (this->right != NULL)
{
delete this->right;
}
}
};

bool sort_by_x(rat_population* first, rat_population* second)
{
return first->x * 10000 + first->y < second->x * 10000 + second->y;
}

bool sort_by_y(rat_population* first, rat_population* second)
{
return first->y * 10000 + first->x < second->y * 10000 + second->x;
}

kdtree_node* build(vector<rat_population*> x_sorted, vector<rat_population*> y_sorted, bool by_x)
{
kdtree_node* result = new kdtree_node();
int num_points = x_sorted.size();
double median_x;
double median_y;
int half = num_points / 2;

if (num_points == 0)
{
result->is_leaf = true;
}
else if (num_points == 1)
{
result->is_leaf = true;
result->data = x_sorted[0];
}
else
{
if (by_x)
{
if (num_points % 2 == 0)
{
median_x = (x_sorted[half - 1]->x + x_sorted[half]->x) / 2.0;
median_y = (x_sorted[half - 1]->y + x_sorted[half]->y) / 2.0;
}
else
{
median_x = x_sorted[half]->x;
median_y = x_sorted[half]->y;
}
result->split_value = median_x;
result->by_x = true;
vector<rat_population*> left_x_sorted;
vector<rat_population*> right_x_sorted;
vector<rat_population*> left_y_sorted;
vector<rat_population*> right_y_sorted;
for (vector<rat_population*>::iterator i = x_sorted.begin(); i != x_sorted.end(); i++)
{
rat_population* current = *i;
if (current->x < median_x)
{
left_x_sorted.push_back(current);
}
else if (current->x == median_x && current->y < median_y)
{
left_x_sorted.push_back(current);
}
else
{
right_x_sorted.push_back(current);
}
}
for (vector<rat_population*>::iterator i = y_sorted.begin(); i != y_sorted.end(); i++)
{
rat_population* current = *i;
if (current->x < median_x)
{
left_y_sorted.push_back(current);
}
else if (current->x == median_x && current->y < median_y)
{
left_y_sorted.push_back(current);
}
else
{
right_y_sorted.push_back(current);
}
}
result->left = build(left_x_sorted, left_y_sorted, false);
result->right = build(right_x_sorted, right_y_sorted, false);
}
else
{
if (num_points % 2 == 0)
{
median_y = (y_sorted[half - 1]->y + y_sorted[half]->y) / 2.0;
median_x = (y_sorted[half - 1]->x + y_sorted[half]->x) / 2.0;
}
else
{
median_y = y_sorted[half]->y;
median_x = y_sorted[half]->x;
}
result->split_value = median_y;
result->by_x = false;
vector<rat_population*> left_x_sorted;
vector<rat_population*> right_x_sorted;
vector<rat_population*> left_y_sorted;
vector<rat_population*> right_y_sorted;
for (vector<rat_population*>::iterator i = x_sorted.begin(); i != x_sorted.end(); i++)
{
rat_population* current = *i;
if (current->y < median_y)
{
left_x_sorted.push_back(current);
}
else if (current->y == median_y && current->x < median_x)
{
left_x_sorted.push_back(current);
}
else
{
right_x_sorted.push_back(current);
}
}
for (vector<rat_population*>::iterator i = y_sorted.begin(); i != y_sorted.end(); i++)
{
rat_population* current = *i;
if (current->y < median_y)
{
left_y_sorted.push_back(current);
}
else if (current->y == median_y && current->x < median_x)
{
left_y_sorted.push_back(current);
}
else
{
right_y_sorted.push_back(current);
}
}
result->left = build(left_x_sorted, left_y_sorted, true);
result->right = build(right_x_sorted, right_y_sorted, true);
}
}

return result;
}

void print_tree(kdtree_node* root, int indent)
{
for (int i = 0; i < indent; i++)
{
cout << " ";
}
if (root->data)
{
cout << "Leaf: (" << root->data->x << ", " << root->data->y << ")" << root->data->size << endl;
}
else
{
if (root->by_x)
{
cout << "Split by X: " << root->split_value << endl;
print_tree(root->left, indent + 1);
print_tree(root->right, indent + 1);
}
else
{
cout << "Split by Y: " << root->split_value << endl;
print_tree(root->left, indent + 1);
print_tree(root->right, indent + 1);
}
}
}

int count(kdtree_node* node)
{
if (node->is_leaf)
{
if (node->data != NULL)
{
return node->data->size;
}
else
{
return 0;
}
}
else
{
return count(node->left) + count(node->right);
}
}

int count(kdtree_node* node, int qMinX, int qMinY, int qMaxX, int qMaxY, int bMinX, int bMinY, int bMaxX, int bMaxY)
{
// Case 0: The node is a leaf node
if (node->is_leaf)
{
if (node->data != NULL)
{
if (qMinX <= node->data->x &&node->data->x <= qMaxX && qMinY <= node->data->y &&node->data->y <= qMaxY)
{
return node->data->size;
}
else
{
return 0;
}
}
else
{
return 0;
}
}

// Case 1: The query region completely includes the bounding box of the current node
if (qMinX <= bMinX && qMinY <= bMinY && qMaxX >= bMaxX && qMaxY >= bMaxY)
{
return count(node);
}

// These value matter only if node->split_value is not an integer
// When split_value is treated as inclusive upper bound, it should be round down (e.g the set of integers less    than or equal to 1.5 is the same set of integer less    than or equals to 1)
// When split_value is treated as inclusive lower bound, it should be round up   (e.g the set of integers greater than or equal to 1.5 is the same set of integer greater than or equals to 2)
int split_lower = (int)node->split_value;
int split_upper = (int)(node->split_value + 0.5);

if (node->by_x)
{
if (qMaxX <= split_lower)
{
return count(node->left, qMinX, qMinY, qMaxX, qMaxY, bMinX, bMinY, split_lower, bMaxY);
}
else if (qMinX >= split_upper)
{
return count(node->right, qMinX, qMinY, qMaxX, qMaxY, split_upper, bMinY, bMaxX, bMaxY);
}
else
{
int leftCount = count(node->left, qMinX, qMinY, qMaxX, qMaxY, bMinX, bMinY, split_lower, bMaxY);
int rightCount = count(node->right, qMinX, qMinY, qMaxX, qMaxY, split_upper, bMinY, bMaxX, bMaxY);
return leftCount + rightCount;
}
}
else
{
if (qMaxY <= split_lower)
{
return count(node->left, qMinX, qMinY, qMaxX, qMaxY, bMinX, bMinY, bMaxX, split_lower);
}
else if (qMinY >= split_upper)
{
return count(node->right, qMinX, qMinY, qMaxX, qMaxY, bMinX, split_upper, bMaxX, bMaxY);
}
else
{
int leftCount = count(node->left, qMinX, qMinY, qMaxX, qMaxY, bMinX, bMinY, bMaxX, split_lower);
int rightCount = count(node->right, qMinX, qMinY, qMaxX, qMaxY, bMinX, split_upper, bMaxX, bMaxY);
return leftCount + rightCount;
}
}

}

int count(kdtree_node* root, int minX, int minY, int maxX, int maxY)
{
return count(root, minX, minY, maxX, maxY, 0, 0, 1024, 1024);
}

int UVa10360()
{
int number_of_cases;
cin >> number_of_cases;
for (int c = 0; c < number_of_cases; c++)
{
int strength;
cin >> strength;
int num_rat_populations;
cin >> num_rat_populations;
list<rat_population*> rat_populations;
for (int p = 0; p < num_rat_populations; p++)
{
rat_population* current_line = new rat_population();
cin >> current_line->x;
cin >> current_line->y;
cin >> current_line->size;
rat_populations.push_back(current_line);
}

// Step 2: Build a spatial index using kd-tree
vector<rat_population*> x_sorted_rat_populations;
vector<rat_population*> y_sorted_rat_populations;

// Step 2.1: Pre-sort the points
for (list<rat_population*>::iterator ri = rat_populations.begin(); ri != rat_populations.end(); ri++)
{
x_sorted_rat_populations.push_back(*ri);
y_sorted_rat_populations.push_back(*ri);
}
sort(x_sorted_rat_populations.begin(), x_sorted_rat_populations.end(), sort_by_x);
sort(y_sorted_rat_populations.begin(), y_sorted_rat_populations.end(), sort_by_y);

// Step 2.2: Recursively build the tree
kdtree_node* root = build(x_sorted_rat_populations, y_sorted_rat_populations, true);
// print_tree(root, 0);

// Step 3: Search for answer
int best_bomb_x = 0;
int best_bomb_y = 0;

map<int, int> result_cache;

for (list<rat_population*>::iterator ri = rat_populations.begin(); ri != rat_populations.end(); ri++)
{
int minBombX = max(strength, (*ri)->x - strength);
int maxBombX = min(1024 - strength, (*ri)->x + strength);
int minBombY = max(strength, (*ri)->y - strength);
int maxBombY = min(1024- strength, (*ri)->y + strength);

for (int bombX = minBombX; bombX <= maxBombX; bombX++)
{
for (int bombY = minBombY; bombY <= maxBombY; bombY++)
{
int bomb_key = bombX * 10000 + bombY;
// A worthwhile check - running the spatial index can be expensive doing repeatedly
// Initializing the array would be expensive enough just for the purpose of checking
if (result_cache.find(bomb_key) == result_cache.end())
{
// Good for verification

/*for (list<rat_population*>::iterator ti = rat_populations.begin(); ti != rat_populations.end(); ti++)
{
if (abs(bombX - (*ti)->x) <= strength && abs(bombY - (*ti)->y) <= strength)
{
}
}*/

dead_rats = count(root, bombX - strength, bombY - strength, bombX + strength, bombY + strength);
// Good for comparison with simple search
// cout << bomb_key << "->" << dead_rats << endl;
bool replace_result = false;
{
replace_result = true;
}
{
if (bombX < best_bomb_x)
{
replace_result = true;
}
else if (bombX == best_bomb_x)
{
if (bombY < best_bomb_y)
{
replace_result = true;
}
}
}
if (replace_result)
{
best_bomb_x = bombX;
best_bomb_y = bombY;
}
}
}
}
}
cout << best_bomb_x << " " << best_bomb_y << " " << max_dead_rats << endl;

// Step 4: Release dynamically allocated memory
for (list<rat_population*>::iterator ri = rat_populations.begin(); ri != rat_populations.end(); ri++)
{
delete (*ri);
}
delete root;
}

return 0;
}

#endif

#ifndef UVa10360_USEKDTREE

int kill[1025][1025];

int UVa10360()
{
int number_of_cases;
cin >> number_of_cases;
for (int c = 0; c < number_of_cases; c++)
{
int strength;
cin >> strength;
int num_rat_populations;
cin >> num_rat_populations;
list<rat_population*> rat_populations;
for (int p = 0; p < num_rat_populations; p++)
{
rat_population* current_line = new rat_population();
cin >> current_line->x;
cin >> current_line->y;
cin >> current_line->size;
rat_populations.push_back(current_line);
}

// Step 2: Build the dead map
for (int i = 0; i < 1025; i++)
{
for (int j = 0; j < 1025; j++)
{
kill[i][j] = 0;
}
}

for (list<rat_population*>::iterator ri = rat_populations.begin(); ri != rat_populations.end(); ri++)
{
int min_bomb_x = max(0,    (*ri)->x - strength);
int min_bomb_y = max(0,    (*ri)->y - strength);
int max_bomb_x = min(1024, (*ri)->x + strength);
int max_bomb_y = min(1024, (*ri)->y + strength);
for (int x = min_bomb_x; x <= max_bomb_x; x++)
{
for (int y = min_bomb_y; y <= max_bomb_y; y++)
{
kill[x][y] += (*ri)->size;
}
}
}

int best_bomb_x = 0;
int best_bomb_y = 0;

// Step 3: Search for solution in the dead map
for (int x = 0; x < 1025; x++)
{
for (int y = 0; y < 1025; y++)
{
{
best_bomb_x = x;
best_bomb_y = y;
}
}
}

cout << best_bomb_x << " " << best_bomb_y << " " << max_dead_rats << endl;

// Step 4: Release dynamically allocated memory
for (list<rat_population*>::iterator ri = rat_populations.begin(); ri != rat_populations.end(); ri++)
{
delete (*ri);
}
}

return 0;
}

#endif

#### 1 comment :

1. Actually a logic you are applying to sort the answer and print the answer with min x and min y is overwhelming. Infact you can calculate the max from i=0, j=0 and update the max only if something is greater than that. This would ensure that if in future there is a x and y with max that is same as max for a smaller x and y, it would not be updated as I have used a > condtion and not a >= condition. Here is the code (AC) in C++:

#include
#include
#include
#include
#include
using namespace std;

int main(){
int t;
cin>>t;
while(t--){
int d;
cin>>d;
int n;
cin>>n;
int a[1025][1025] = {};
for(int i=0; i>x>>y>>p;
for(int j=max(0, x-d); j<=min(x+d, 1024); j++){
for(int k=max(0, y-d); k<=min(y+d, 1024); k++){
a[j][k]+=p;
}
}
}
int max = -1;
int maxX = -1;
int maxY = -1;
for(int i=0; i<1025; i++){
for(int j=0; j<1025; j++){
if (a[i][j]>max){
max = a[i][j];
maxX = i;
maxY = j;
}
}
}
cout << maxX << " " << maxY << " " << max << endl;
}

}

By The way, Thanx, you helped me figure out the problem of min(x+d, 1024) as I was missing that.