USACO Camelot解題レポート

この問題もネット上の解題レポートを見て出した解法です...
まず,knightがkingを受け取る点は碁盤全体で列挙する必要はなく,kingは最大2ステップmoveである.これはすべてのテストポイントを通過することができます.なぜかというと、ネット上でも証明書を探したことがあるが、見る気にならなかった.
次にfloydは非常に非効率なアルゴリズムであり,n^3を要する時間的複雑さを決定するため,密集図(アルゴリズムはエッジの個数に関係なく)に適しているが,我々が遭遇するのは一般的に疎図である.SPFAもBFSももっといいです.以前johnsonアルゴリズムは疎図に適しているようだ.
再び、usacoがくれたスレッドが読めませんでした..

/*
ID: thestor1
LANG: C++
TASK: camelot
*/
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <climits>
#include <vector>
#include <cassert>
#include <string>
#include <algorithm>
#include <stack>
#include <set>
#include <queue>

using namespace std;

struct Position{
	int r, c;

	Position()
	{
	}

	Position(int r, int c)
	{
		this->r = r;
		this->c = c;
	}

	Position(const Position& other):r(other.r), c(other.c){

	}

	Position& operator =(const Position& other){
		r=other.r;
		c=other.c;
		return *this;
	}
};

const int MAXR = 26;
const int MAXC = 30;
const int MAXN = MAXR * MAXC;

int dis[MAXR][MAXC][MAXR][MAXC];

const int directx[] = {-2, -1, 1, 2, 2, 1, -1, -2};
const int directy[] = {1, 2, 2, 1, -1, -2, -2, -1};

bool marked[MAXR][MAXC];
int index(int r, int c, int C)
{
	return r * C + c;
}

bool isWithin(int r, int c, int R, int C)
{
	return r >= 0 && r < R && c >= 0 && c < C;
}

void spfa(const Position &s, const int R, const int C)
{
	if(marked[s.r][s.c])
	{
		return;
	}
	marked[s.r][s.c] = true;
	for(int i = 0; i < R; ++i)
	{
		for(int j = 0; j < C; ++j)
		{
			dis[s.r][s.c][i][j] = INT_MAX;
		}
	}
	dis[s.r][s.c][s.r][s.c] = 0;

	deque<Position> que;
	bool onQueue[MAXR][MAXC];
	memset(onQueue, false, MAXN * sizeof(bool));

	//int s = index(r, c, C);
	que.push_back(s);
	onQueue[s.r][s.c] = true;

	while(!que.empty())
	{
		/*
		if(que.size() == 51 && que.front().r == 24 && que.front().c == 18)
		{
		fprintf(stdout, "debug
");
		}
		fprintf(stdout, "que.size: %d
", que.size());
		*/
		Position u = que.front();
		que.pop_front();
		//fprintf(stdout, "u: (%d, %d)
", u.r, u.c);
		if(dis[s.r][s.c][u.r][u.c] == INT_MAX)
		{
			continue;
		}
		onQueue[u.r][u.c] = false;
		for(int d = 0; d < 8; ++d)
		{
			int vr = u.r + directx[d];
			int vc = u.c + directy[d];
			if(!isWithin(vr, vc, R, C))
			{
				continue;
			}

			Position v(vr, vc);
			//v.r = vr;
			//v.c = vc;
			if(dis[s.r][s.c][u.r][u.c] + 1 < dis[s.r][s.c][v.r][v.c])
			{
				dis[s.r][s.c][v.r][v.c] = dis[s.r][s.c][u.r][u.c] + 1;
				if(!onQueue[v.r][v.c])
				{
					if(que.empty() || 
						dis[s.r][s.c][v.r][v.c] < dis[s.r][s.c][que.front().r][que.front().c])
					{
						que.push_front(v);
					}
					else
					{
						que.push_back(v);
					}
					onQueue[v.r][v.c] = true;
				}
			}
		}
	}
}


int main()
{
	FILE *fin  = fopen ("camelot.in", "r");
	FILE *fout = fopen ("camelot.out", "w");
	//freopen("log.txt", "w", stdout);

	int R, C;
	fscanf(fin, "%d%d
", &C, &R);
	//int N = R * C;
	
	memset(marked, false, R * C * sizeof(bool));

	Position king;
	char r;
	fscanf(fin, "%c", &r);
	king.r = r - 'A';
	int c;
	fscanf(fin, "%d
", &c);
	king.c = c - 1;
	assert(king.r < R && king.c < C);
	//fprintf(stdout, "king: 
%d %d
", king.r, king.c);
	vector<Position> knights;
	while(fscanf(fin, "%c", &r) > 0)
	{
		if(r < 'A' || r > 'Z')
		{
			continue;
		}
		Position knight;
		knight.r = r - 'A';
		fscanf(fin, " %d", &c);
		knight.c = c - 1;
		knights.push_back(knight);
	}

	//for(int i = 0; i < knights.size(); ++i)
	//{
	//assert(knights[i].r < R && knights[i].c < C);
	//fprintf(stdout, "%c %d
", knights[i].r + 'A', knights[i].c + 1);
	//}
	//fprintf(stdout, "number of knights: %d
", knights.size());

	for(int i = 0; i < knights.size(); ++i)
	{
		spfa(knights[i], R, C);
		/*
		for(int r = 0; r < R; ++r)
		{
		for(int c = 0; c < C; ++c)
		{
		fprintf(stdout, "%d\t", dis[knights[i].r][knights[i].c][r][c]);
		}
		fprintf(stdout, "
");
		}
		*/
	}

	int minmoves = -1;
	for(int gr = 0; gr < R; ++gr)//gathering point
	{
		for(int gc = 0; gc < C; ++gc)
		{
			int knightsdis = 0;
			for(int i = 0; i < knights.size(); ++i)
			{
				knightsdis += dis[knights[i].r][knights[i].c][gr][gc];
			}
			int kingmove = 0;
			int pr, pc;
			int knightsmove = 0;
			for(int kingmover = -2; kingmover <= 2; ++kingmover)
			{
				for(int kingmovec = -2; kingmovec <= 2; ++kingmovec)
				{
					kingmove = max(abs(kingmover), abs(kingmovec));
					//if(!kingmover && !kingmovec)
					//{
						//kingmove = 0;
					//}
					//else
					//{
						//kingmove = 1;
					//}
					pr = king.r + kingmover;
					pc = king.c + kingmovec;
					if(!isWithin(pr, pc, R, C))
					{
						continue;
					}
					spfa(Position(pr, pc), R, C);
					for(int pk = 0; pk < knights.size(); ++pk) //pickup knight
					{
						if(dis[knights[pk].r][knights[pk].c][pr][pc] == INT_MAX 
							|| dis[pr][pc][gr][gc] == INT_MAX)
						{
							continue;
						}
						knightsmove = knightsdis - dis[knights[pk].r][knights[pk].c][gr][gc] + dis[knights[pk].r][knights[pk].c][pr][pc] + dis[pr][pc][gr][gc];
						int moves = knightsmove + kingmove;
						if(minmoves < 0 || moves < minmoves)
						{
							minmoves = moves;
						}
					}
				}
			}
		}
	}
	fprintf(fout, "%d
", minmoves < 0 ? 0 : minmoves);
	return 0;
}