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 <cstdio>#include <string>#include <algorithm>#include <cstring>#include <iostream>#include <cmath>#include <cctype>#include <vector>#include <set>#include <queue>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<class T>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<class T>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<class T>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<node> 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;}