题目相关

题目链接

AtCoder Beginner Contest 184 E 题，https://atcoder.jp/contests/abc184/tasks/abc184_e。

Problem Statement

There is a town represented as a two-dimensional grid with H horizontal rows and W vertical columns.
A character ai,j describes the square at the i-th row from the top and j-th column from the left. Here, ai,j is one of the following: S , G , . , # , a, ..., and z.
# represents a square that cannot be entered, and a, ..., z represent squares with teleporters.

Takahashi is initially at the square represented as S. In each second, he will make one of the following moves:

Go to a non-# square that is horizontally or vertically adjacent to his current position.
Choose a square with the same character as that of his current position, and teleport to that square. He can only use this move when he is at a square represented as a, ..., or z.

Find the shortest time Takahashi needs to reach the square represented as G from the one represented as S.
If the destination is unreachable, report -1 instead.

Input

Input is given from Standard Input in the following format:

H W
a1,1 ... a1,w
.
.
.
ah,1 ... ah,w

Output

Print the shortest time Takahashi needs to reach the square represented as G from the one represented as S.
If the destination is unreachable from the initial position, print -1 instead.

Samples1

Sample Input 1

2 5
S.b.b
a.a.G

Sample Output 1

Explaination

Let (i,j) denote the square at the i-th row from the top and j-th column from the left.
Initially, Takahashi is at (1,1). One way to reach (2,5) in four seconds is:

go from (1,1) to (2,1);
teleport from (2,1) to (2,3), which is also an a square;
go from (2,3) to (2,4);
go from (2,4) to (2,5).

Samples2

Sample Input 2

11 11
S##...#c...
...#d.#.#..
..........#
.#....#...#
#.....bc...
#.##......#
.......c..#
..#........
a..........
d..#...a...
.#........G

Sample Output 2

Samples3

Sample Input 3

11 11
.#.#.e#a...
.b..##..#..
#....#.#..#
.#dd..#..#.
....#...#e.
c#.#a....#.
.....#..#.e
.#....#b.#.
.#...#..#..
......#c#G.
#..S...#...

Sample Output 3

-1

Constraints

1≤H,W≤2000
ai,j is S, G, ., #, or a lowercase English letter. There is exactly one square represented as S and one square represented as G.

题解报告

题目翻译

有一个城镇可以用二维网格表示，该城镇水平方向有 H 行，垂直方向有 W 列。字符 $a_{i,j}$ 描述最上边开始的第 i 行最左面开始的第 j 列，它的值是：S，G，.，#，a，...，z 其中一个。# 表示这个点不能进入，小写字母 a ~ z 表示传送点。

高桥开始的的位置为 S，他每次可以按照以下规则移动：

向相邻的位置移动，只要这个位置不是 #。
如果这里是传送点，高桥可以传送到字母相同的传送点。

请找出高桥从 S 到 G 的最短路径，如果不存在，输出 -1。

题目分析

看完题目描述，走迷宫并且是找出最短路线，那么必然使用 BFS。本题和 BFS 模板相比，多了一个传送点，如何解决传送点问题。根据题目意思，我们一共有四种走法，即上、右、下、左。

我的思路是这样的：

1、在某个点判断下一个位置的时候，不考虑下一个点是否是传送点，和原有的 BFS 一样进行移动。

2、所有下一个可能位置计算完成后，判断这个点是否是传送点。如果是传送点，遍历修改传送点的代价 cost。修改结束，清空所有传送点。

下面我们使用样例数据来验证一下思路。

样例数据 1 分析

根据样例数据，起点在 (1,1) 位置。一般算法竞赛都将左上角定义为 1,1。起点的代价 cost 为 0，其余点的 cost 都为 INT_MAX。我们按照上、右、下、左进行遍历。首先将起点 (1,1) 加入到 BFS 的遍历队列中。下面我们开始模拟整个 BFS 遍历过程。

初始状态

cost 数组的值为

	1	2	3	4	5
1	0	INT_MAX	INT_MAX	INT_MAX	INT_MAX
2	INT_MAX	INT_MAX	INT_MAX	INT_MAX	INT_MAX

	1	2	3	4	5
1	0	1	INT_MAX	INT_MAX	INT_MAX
2	1	INT_MAX	INT_MAX	INT_MAX	INT_MAX

	1	2	3	4	5
1	0	1	INT_MAX	INT_MAX	INT_MAX
2	1	2	2	INT_MAX	INT_MAX

	1	2	3	4	5
1	0	1	2	INT_MAX	INT_MAX
2	1	2	2	INT_MAX	INT_MAX

	1	2	3	4	5
1	0	1	2	INT_MAX	INT_MAX
2	1	2	2	INT_MAX	INT_MAX

	1	2	3	4	5
1	0	1	2	INT_MAX	INT_MAX
2	1	2	2	3	INT_MAX

	1	2	3	4	5
1	0	1	2	3	3
2	1	2	2	3	INT_MAX

	1	2	3	4	5
1	0	1	2	3	3
2	1	2	2	3	4

	1	2	3	4	5
1	0	1	2	3	3
2	1	2	2	3	4

	1	2	3	4	5
1	0	1	2	3	3
2	1	2	2	3	4

	1	2	3	4	5
1	0	1	2	3	3
2	1	2	2	3	4

到这里为止，BFS 队列为空，说明我们遍历了迷宫的所有点。遍历完成。

注意：只要 BFS 第一次走到终点，BFS 就可以停止了，因为 BFS 可以保证第一次就是最短路径。

AC 参考代码

//https://atcoder.jp/contests/abc184/tasks/abc184_e
//E - Third Avenue
#include <bits/stdc++.h>using namespace std;//如果提交到OJ，不要定义 __LOCAL
//#define __LOCALconst int MAXH=2e3+4;
const int MAXW=2e3+4;
char maze[MAXH][MAXW];
int cost[MAXH][MAXW];int h,w;
pair<int, int> qd;
pair<int, int> zd;
vector<pair<int, int>> tele[26];const int dx[]={0, 1, 0, -1};
const int dy[]={-1, 0, 1, 0};int bfs() {queue<pair<int, int>> q;q.push(qd);cost[qd.first][qd.second]=0;while (true!=q.empty()) {pair<int, int> cur = q.front();q.pop();int nc=cost[cur.first][cur.second]+1;//下一个位置的代价//新位置for (int i=0; i<4; i++) {int nx = cur.first+dx[i];int ny = cur.second+dy[i];//合法性判断if (nx<1||nx>h||ny<1||ny>w||'#'==maze[nx][ny]) {continue;}//路径判断if (nc<cost[nx][ny]) {cost[nx][ny]=nc;q.emplace(nx, ny);//判断是不是终点if (nx==zd.first && ny==zd.second) {return nc;}}}//如果是传送点if (maze[cur.first][cur.second]>='a' && maze[cur.first][cur.second]<='z') {//将所有的vector<pair<int, int>> &t=tele[maze[cur.first][cur.second]-'a'];vector<pair<int, int>>::iterator it;for (it=t.begin(); it!=t.end(); it++) {if (nc<cost[it->first][it->second]) {cost[it->first][it->second]=nc;q.emplace(it->first, it->second);}}t.clear();}}return -1;
}int main() {
#ifndef __LOCAL//这部分代码需要提交到OJ，本地调试不使用ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#endifcin>>h>>w;for (int i=1; i<=h; i++) {for (int j=1; j<=w; j++) {cost[i][j]=INT_MAX;cin>>maze[i][j];if ('S'==maze[i][j]) {//起点qd.first=i;qd.second=j;cost[i][j]=0;} else if ('G'==maze[i][j]) {zd.first=i;zd.second=j;} else if (maze[i][j]>='a' && maze[i][j]<='z') {tele[maze[i][j]-'a'].emplace_back(i, j);}}}//开始BFScout<<bfs()<<"\n";#ifdef __LOCAL//这部分代码不需要提交到OJ，本地调试使用system("pause");
#endifreturn 0;
}

时间复杂度

O(HW)。

空间复杂度