HDU 2579 - Dating with girls(2)

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传送门HDU 2579 - Dating with girls(2)

看到这题我就想到了UVa 10047 - The Monocycle

只是上一题更复杂一点。

也是用优先队列。

这时候的状态比一般的状态还要多一个，即秒数，所以要开一个三维数组。

A了之后看了很多解题报告，都是直接用队列，但是个人感觉不太靠谱。原因和上面提到的那题一样。

#include <cstdio>#include <cstring>#include <queue>using namespace std;const int dir[][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};struct POINT{friend bool operator< (const POINT &a, const POINT &b){return a.sec > b.sec;}int x, y;int sec;};priority_queue<POINT> qu;POINT start, target;char maze[110][110];int ans, row, col, k, vis[110][110][12];bool BFS();int main(){//freopen("input.txt", "r", stdin);int i, j, T;scanf("%d", &T);while (T--){memset(vis, 0, sizeof(vis));scanf("%d%d%d%*c", &row, &col, &k);for (i = 0; i < row; i++)gets(maze[i]);for (i = 0; i < row; i++)for (j = 0; j < col; j++){if (maze[i][j] == 'Y')start.x = i, start.y = j;if (maze[i][j] == 'G')target.x = i, target.y = j;}bool flag = BFS();if (!flag)printf("Please give me another chance!\n");elseprintf("%d\n", ans);}return 0;}bool BFS(){while (!qu.empty())qu.pop();start.sec = 0;qu.push(start);vis[start.x][start.y][0] = 1;while (!qu.empty()){POINT cur = qu.top();qu.pop();for (int i = 0; i < 4; i++){POINT temp;bool inFlag = false;int dx = cur.x + dir[i][0];int dy = cur.y + dir[i][1];int ds = cur.sec + 1;if (dx >= 0 && dx < row && dy >= 0 && dy < col)//如果没越界{temp.x = dx, temp.y = dy, temp.sec = ds;if (maze[dx][dy] == '#')//如果是‘#’，只有在倍数的情况下,且没有被访问过,可以进。{if (ds % k == 0){if (!vis[dx][dy][ds % k]){inFlag = true;qu.push(temp);vis[dx][dy][ds % k] = 1;}}}else//如果不是墙if (!vis[dx][dy][ds % k])//if three states are different, push into the queue{inFlag = true;qu.push(temp);vis[dx][dy][ds % k] = 1;}if (inFlag)if (temp.x == target.x && temp.y == target.y){ans = temp.sec;return true;}}}}return false;}

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