UVA 10798 - Be wary of Roses（记忆化BFS）

UVA 10798 - Be wary of Roses

题目链接

题意：给定一个地图，人一开始在中心，问选择一种走法走出去，使得面朝任何一个方向走，踩到的花的最大值最小

思路：用优先队列进行BFS，每次取出踩到最少的情况，广搜记录状态为当前位置，和4个方向分别踩到的花数

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int N = 21;
const int d[4][2] = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};

int n, vis

[11][11][11][11];

char g

;

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));
}
}
}
}

int main() {
while (~scanf("%d", &n) && n) {
init();
printf("At most %d rose(s) trampled.\n", bfs());
}
return 0;
}