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

char g[N][N];

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;
}

UVA 10798 - Be wary of Roses（记忆化BFS）

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原文地址：http://blog.csdn.net/accelerator_/article/details/38901669