省选模拟（12.10） T3 stwell

STWELL

题目背景：

12.10 省选模拟T3

分析：BFS

 

注意到我们可以过来做会更简单一些。我们把每个点拆成四个点, 分别表示这个点是从哪里转移来的, 然后我们要求到达每个点所需

的能量最小值，如果我们当前在(x, y), 从pos转移来, 那么我们有两种转移

1. 转移到(x + dx[pos], y + dy[pos], pos), 更新答案为CurK + 1

2. 如果(x, y) 处无山, 则分别向四个方向距离为CurK的点转移, 更新答案为CurK

注意到第一种转移因为记录了从哪转移来, 所以显然是正确的, 我们也可以证明这些转移就是充分必要的

 

Source:

/*
	created by scarlyw
*/
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

inline char read() {
	static const int IN_LEN = 1024 * 1024;
	static char buf[IN_LEN], *s, *t;
	if (s == t) {
		t = (s = buf) + fread(buf, 1, IN_LEN, stdin);
		if (s == t) return -1;
	}
	return *s++;
}

///*
template
inline void R(T &x) {
	static char c;
	static bool iosig;
	for (c = read(), iosig = false; !isdigit(c); c = read()) {
		if (c == -1) return ;
		if (c == '-') iosig = true;	
	}
	for (x = 0; isdigit(c); c = read()) 
		x = ((x << 2) + x << 1) + (c ^ '0');
	if (iosig) x = -x;
}
//*/

const int OUT_LEN = 1024 * 1024;
char obuf[OUT_LEN], *oh = obuf;
inline void write_char(char c) {
	if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;
	*oh++ = c;
}


template
inline void W(T x) {
	static int buf[30], cnt;
	if (x == 0) write_char('0');
	else {
		if (x < 0) write_char('-'), x = -x;
		for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
		while (cnt) write_char(buf[cnt--]);
	}
}

inline void flush() {
	fwrite(obuf, 1, oh - obuf, stdout);
}

/*
template
inline void R(T &x) {
	static char c;
	static bool iosig;
	for (c = getchar(), iosig = false; !isdigit(c); c = getchar())
		if (c == '-') iosig = true;	
	for (x = 0; isdigit(c); c = getchar()) 
		x = ((x << 2) + x << 1) + (c ^ '0');
	if (iosig) x = -x;
}
//*/

const int MAXN = 2000 + 10;

const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};

int n, m, k, sx, sy, tx, ty;
int d[MAXN][MAXN][4], map[MAXN][MAXN];

struct node {
	int x, y, pos;
	node(int x = 0, int y = 0, int pos = 0) : x(x), y(y), pos(pos) {};
} ;

inline void read_in() {
	R(n), R(m), R(k), R(sx), R(sy), R(tx), R(ty);
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < m; ++j)
			R(map[i][j]);
}

std::deque q;
inline void update(int x, int y, int pos, int w) {
	if (d[x][y][pos] > w) {
		d[x][y][pos] = w;
		if (q.empty()) {
			q.push_back(node(x, y, pos));
			return ;
		}
		node e = q.front();
		if (d[x][y][pos] <= d[e.x][e.y][e.pos]) q.push_front(node(x, y, pos));
		else q.push_back(node(x, y, pos)); 
	}
}

inline void solve() {
	memset(d, 127, sizeof(d));
	for (int i = 0; i < 4; ++i) update(tx, ty, i, 0);
	while (!q.empty()) {
		node e = q.front();
		q.pop_front();
		int w = d[e.x][e.y][e.pos];
		if (e.x == sx && e.y == sy && w <= k) std::cout << "Possible", exit(0);
		int x = e.x + dx[e.pos], y = e.y + dy[e.pos];
		if (x >= 0 && y >= 0 && x < n && y < m) update(x, y, e.pos, w + 1);
		if (map[e.x][e.y] == 1) continue ;
		for (int i = 0; i < 4; ++i) {
			int x = e.x + dx[i] * w, y = e.y + dy[i] * w;
			if (x >= 0 && y >= 0 && x < n && y < m) update(x, y, i, w);
		}
	}
	std::cout << "Impossible", exit(0);
}

int main() {
	freopen("stwell.in", "r", stdin);
	freopen("stwell.out", "w", stdout); 
	read_in();
	solve();
	return 0;
}