UVA 10798 - Be wary of Roses （bfs+hash）

参考http://blog.csdn.net/tobewhatyouwanttobe/article/details/17403987和http://blog.csdn.net/accelerator_/article/details/38901669

output

For each case, output a line containing the minimum guaranteed number of roses you can step on while escaping.

5

.RRR.

R.R.R

R.P.R

R.R.R

.RRR.

0

At most 2 rose(s) trampled.

题意：

UVA的题就是不好读懂呀，我会说我搞了两个小时才看懂题意么?

意思就是 让你确定一条方案 这个方案保证无论你开始面朝哪个方向，用这个方案走出去后使得踩到的玫瑰最小。如样例（上、上、上 这个方案就能保证你开始无论朝那个方向，这样走踩到的玫瑰最小为2），方便理解，解释一下我贴的第一组样例，这个样例 就可以确定 “上、上、左、左、左、左、左 ” 这个方案使得答案最优为 3。

要点：看到能够存的下，就存，，然后用旋转的原理；、

if (g[xx][yy] == 'R') up++;
if (g[n - 1 - yy][xx] == 'R') left++;
if (g[n - 1 - xx][n - 1 - yy] == 'R') down++;
if (g[yy][n - 1 - xx] == 'R') right++;

struct State {
int x, y, val;
int up, left, down, right;
State() {x = y = up = left = down = right = 0;}
State(int x, int y, int up, int left, int down, int right) {
this->x = x;
this->y = y;
this->up = up;
this->left = left;
this->down = down;
this->right = right;
val = max(max(max(up,left), down), right);
}
bool operator < (const State& c) const {
return val > c.val;
}
} s;

void init() {
for (int i = 0; i < n; i++) {
scanf("%s", g[i]);
for (int j = 0; j < n; j++)
if (g[i][j] == 'P')
s.x = i, s.y = j;
}
}

int bfs() {
memset(vis, 0, sizeof(vis));
priority_queue<State> Q;
Q.push(s);
vis[s.x][s.y][0][0][0][0] = 1;
while (!Q.empty()) {
State u = Q.top();
Q.pop();
if (u.x == 0 || u.x == n - 1 || u.y == 0 || u.y == n - 1) return u.val;
for (int i = 0; i < 4; i++) {
int xx = u.x + d[i][0];
int yy = u.y + d[i][1];
int up = u.up;
int left = u.left;
int down = u.down;
int right = u.right;
if (g[xx][yy] == 'R') up++;
if (g[n - 1 - yy][xx] == 'R') left++;
if (g[n - 1 - xx][n - 1 - yy] == 'R') down++;
if (g[yy][n - 1 - xx] == 'R') right++;
if (!vis[xx][yy][up][left][down][right]) {
vis[xx][yy][up][left][down][right] = 1;
Q.push(State(xx, yy, up, left, down, right));
}
}
}
}