HexaBlitz

The fun and addictive match-three / bubble-shooter game revisited for Android.

WHAT’S HEXABLITZ?

The idea of this implementation is to convert the game from a strategy/action-planning game into a fast-paced survival aimed at competing for high scores, breaking your way through unlimited levels, and sharing the experience with others. The game is no longer turn based, you don’t control your cannon anymore; It charges and fires automatically in given intervals. Your challenge is to adjust the aiming angle correctly before it fires.

GAMEPLAY

You level up after firing some amount of hexagons which are indicated by the bottom bar.
The game’s pace speeds up with each level and you’re presented with new types of hexagons each set of levels.
As usual, matching three bubbles of the same color earns some points.
Moreover, collecting more than three gives you a super hexagon. For different colors, each super hexagon has a unique effect.

 

DOWNLOAD

• DropBox – HexaBlitz0.9

CONTESTS

• AAC2013 — First place Egypt. Top 10 in final round.

10130 – SuperSale

A 0/1-Knapsack problem.
Apply the 0/1-Knapsack for each family member given the weight they can carry and sum the maximum up.
I added a little detail but it doesn’t make much of a difference. I sort the items according to their weights and start from light items until I reach an item I can’t add. Items heavier than that last item are, of course, no use so we return from there and stop the recursion.
C++ implementation:

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

#define N 1004
#define G 104
#define W 34
#define weight first
#define price second

int dp[N][W];
pair<int, int> item[N];

bool cmp(pair<int, int> p1, pair<int, int> p2) {
	return p1.weight > p2.weight;
}

int getMax(int n, int w) {
	if (n < 0)
		return 0;
	if (dp[n][w] != -1)
		return dp[n][w];
	if (w < item[n].weight)
		return dp[n][w] = 0;
	else
		return dp[n][w] = max(getMax(n - 1, w - item[n].weight) + item[n].price, getMax(n - 1, w));
}

int main() {
	int totT, totN, totG;
	cin >> totT;
	for (int i = 0; i < totT; ++i) {
		// input
		cin >> totN;
		for (int j = 0; j < totN; ++j)
			cin >> item[j].price >> item[j].weight;
		memset(dp, -1, sizeof dp);
		sort(item, item + totN, cmp);
		cin >> totG;
		// solve
		int sum = 0;
		for (int j = 0, w; j < totG; ++j) {
			cin >> w;
			sum += getMax(totN - 1, w);
		}
		cout << sum << endl;
	}
	return 0;
}