Curling 2.0
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 30740 Accepted: 12179Description
On Planet MM-21, after their Olympic games this year, curling is getting popular. But the rules are somewhat different from ours. The game is played on an ice game board on which a square mesh is marked. They use only a single stone. The purpose of the game is to lead the stone from the start to the goal with the minimum number of moves.
Fig. 1 shows an example of a game board. Some squares may be occupied with blocks. There are two special squares namely the start and the goal, which are not occupied with blocks. (These two squares are distinct.) Once the stone begins to move, it will proceed until it hits a block. In order to bring the stone to the goal, you may have to stop the stone by hitting it against a block, and throw again.
Fig. 1: Example of board (S: start, G: goal)
The movement of the stone obeys the following rules:
At the beginning, the stone stands still at the start square.The movements of the stone are restricted to x and y directions. Diagonal moves are prohibited.When the stone stands still, you can make it moving by throwing it. You may throw it to any direction unless it is blocked immediately(Fig. 2(a)).Once thrown, the stone keeps moving to the same direction until one of the following occurs: The stone hits a block (Fig. 2(b), (c)). The stone stops at the square next to the block it hit.The block disappears.The stone gets out of the board. The game ends in failure.The stone reaches the goal square. The stone stops there and the game ends in success.You cannot throw the stone more than 10 times in a game. If the stone does not reach the goal in 10 moves, the game ends in failure.Fig. 2: Stone movements
Under the rules, we would like to know whether the stone at the start can reach the goal and, if yes, the minimum number of moves required.
With the initial configuration shown in Fig. 1, 4 moves are required to bring the stone from the start to the goal. The route is shown in Fig. 3(a). Notice when the stone reaches the goal, the board configuration has changed as in Fig. 3(b).
Fig. 3: The solution for Fig. D-1 and the final board configuration
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. The number of datasets never exceeds 100.
Each dataset is formatted as follows.
the width(=w) and the height(=h) of the board First row of the board ... h-th row of the board
The width and the height of the board satisfy: 2 <= w <= 20, 1 <= h <= 20.
Each line consists of w decimal numbers delimited by a space. The number describes the status of the corresponding square.
0vacant square1block2start position3goal positionThe dataset for Fig. D-1 is as follows:
6 6 1 0 0 2 1 0 1 1 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 1 0 0 0 0 1 0 1 1 1 1 1
Output
For each dataset, print a line having a decimal integer indicating the minimum number of moves along a route from the start to the goal. If there are no such routes, print -1 instead. Each line should not have any character other than this number.
Sample Input
2 1 3 2 6 6 1 0 0 2 1 0 1 1 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 1 0 0 0 0 1 0 1 1 1 1 1 6 1 1 1 2 1 1 3 6 1 1 0 2 1 1 3 12 1 2 0 1 1 1 1 1 1 1 1 1 3 13 1 2 0 1 1 1 1 1 1 1 1 1 1 3 0 0Sample Output
1 4 -1 4 10 -1题目大意:蛮有趣的一题，大概就是一个小球在冰面上滑并且给出起始点和终点，但每次move只能沿x或y方向，冰面上还有若干block,每次小球撞到block会停下来，并且block会被击碎，注意每改变一次小球的方向，小球会一直朝这个方向运动，直到撞到block或者出界或者到达终点。出界或者move超过10次均算失败，问经过几次move小球可以到达终点。
分析：题中给出的数据非常小，长宽均不超过20，所以深搜就可解决，搜索出所有状态并进行更新就可得出答案，此题的关键在于每次状态的更新，因为本题和一般的深搜题不一样，小球会沿着一个方向一直滑，而不是每次只移动一格。
注意事项：注意剪枝，否则会TLE。
代码：
#include <cstdio> const int maxn = 25; int N, M, res; int a[maxn][maxn]; int dx[4] = { 0,1,0,-1 }; int dy[4] = { -1,0,1,0 }; void dfs(int x, int y, int temp) { temp++; if (temp > res || temp > 10) return; //剪枝，不然会TLE for (int i = 0; i < 4; i++) { int nx = x, ny = y; int tag = 0; while (true) { nx += dx[i]; ny += dy[i]; if (nx >= 0 && nx < N&&ny >= 0 && ny < M && (a[nx][ny] == 0 || a[nx][ny] == 2)) { tag = 1; continue; } else if (nx >= 0 && nx < N&&ny >= 0 && ny < M&&a[nx][ny] == 1) { a[nx][ny] = 0; if(tag) dfs(nx - dx[i], ny - dy[i], temp); a[nx][ny] = 1; break; } else if (nx >= 0 && nx < N&&ny >= 0 && ny < M&&a[nx][ny] == 3) { if (temp < res) { res = temp; } break; } else break; } } return; } int main() { int sx, sy; while (true) { res = 100000000; scanf("%d %d", &M, &N); if (N == 0 && M == 0) break; for (int i = 0; i < N; i++) { for (int j = 0; j < M; j++) { scanf("%d", &a[i][j]); if (a[i][j] == 2) { sx = i; sy = j; } } } dfs(sx, sy, 0); if (res > 10) printf("-1\n"); else printf("%d\n", res); } return 0; }
