SRM 564

呜呜呜呜呜,被虐爆了……

第一题就不知道怎么做,后来自己归纳了一下才发现象棋的规律,还因为写错了一个地方被challenge了,呜呜呜呜,我简直弱爆了。

第二题组合数学加动态规划,第三题估计也是动态规划,完全没有思路。。。。

我还是滚去好好学我的离散数学吧。。。。

 

division2的第三题真是好题,让我下午做了整整3个小时,最后卡时过了(标程应该也超时吧……)

分析可见http://apps.topcoder.com/wiki/display/tc/SRM+564,非常非常精彩,强烈推荐。

我的代码:

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <string.h>
#include <ctime>

using namespace std;

class KnightCircuit {
public:
	long long maxSize(int, int, int, int);
	int gcd(int a, int b);
	long long bfs(int w, int h, int a, int b);
};

//when the scale is very small, just use breath first search
long long KnightCircuit::bfs(int w, int h, int a, int b){
	bool** check = new bool*[h+1];
	for(int i=0; i <= h; i++)
		check[i] = new bool[w+1];	
	//mark the corresponding directions
	int dir_x[]={a, a, -a, -a, b, b, -b, -b};
	int dir_y[]={b, -b, b, -b, a, -a, a, -a};
	long long goal=0;
	
	clock_t start=clock();
	for(int m=0; m<h; m++)
	for(int n=0; n<w; n++){		
		clock_t end=clock();
		if((double)(end-start)/CLOCKS_PER_SEC > 1.0)
			return goal;
		for(int i=0; i<=h; i++)
		for(int j=0; j<=w; j++)
			check[i][j] = false;
		check[m][n] = true;
		queue<int> q;
		while(q.empty() == false)
			q.pop();
		q.push(m); q.push(n);
		long long res=1;
		while(q.empty() == false){
			int x = q.front();	q.pop();
			int y = q.front();	q.pop();
			//search for the next step
			for(int i=0; i<8; i++){
				int tmp_x = x + dir_x[i];
				int tmp_y = y + dir_y[i];
				if(tmp_x < 0 || tmp_y < 0 || tmp_x >= h || tmp_y >= w)
					continue;
				if(check[tmp_x][tmp_y] == false){
					check[tmp_x][tmp_y] = true;
					q.push(tmp_x);
					q.push(tmp_y);
					res++;
				}//end if
			}//end for loop
		}//end while loop	
		goal=max(goal, res);
	}
	
	return goal;
}//end method bfs

//hunt for the greatest common divisor using Euclid method
int KnightCircuit::gcd(int a, int b){
    while(a > 0){
		int c = a;
		a = b % a;
		b = c;    
    }
    return b;
}

long long KnightCircuit::maxSize(int w, int h, int a, int b) {
	int g = gcd(a, b);
	//minimize the scale of problem
	if( g != 1)
		return maxSize((1 + (w - 1)/g), (1 + (h - 1)/g), a/g, b/g);
	//just use breath first search	
	if((w <= 20) || (h <= 20))
		return bfs(w, h, a, b);
	//all positions are reachable
	if((a + b) % 2 == 1)
		return w * (long long)h;
	//only half positions are reachable
	else
		return (1 + (long long)w * h)/2;
}

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