题目相关

题目链接

AtCoder Beginner Contest 170 F题，https://atcoder.jp/contests/abc170/tasks/abc170_f。

Problem Statement

Snuke, a water strider, lives in a rectangular pond that can be seen as a grid with H east-west rows and W north-south columns. Let (i,j) be the square at the i-th row from the north and j-th column from the west.

Some of the squares have a lotus leaf on it and cannot be entered. The square (i,j) has a lotus leaf on it if cij is @, and it does not if cij is ..

In one stroke, Snuke can move between 1 and K squares (inclusive) toward one of the four directions: north, east, south, and west. The move may not pass through a square with a lotus leaf. Moving to such a square or out of the pond is also forbidden.

Find the minimum number of strokes Snuke takes to travel from the square (x1,y1) to (x2,y2). If the travel from (x1,y1) to (x2,y2) is impossible, point out that fact.

Input

Input is given from Standard Input in the following format:

H W K
x1 y1 x2 y2
c1,1 c1,2 .. c1,W
c2,1 c2,2 .. c2,W
.
.
cH,1 cH,2 .. cH,W

Output

Print the minimum number of strokes Snuke takes to travel from the square (x1,y1) to (x2,y2), or print -1 if the travel is impossible.

Samples1

Sample Input 1

3 5 2
3 2 3 4
.....
.@..@
..@..

Sample Output 1

Explaination

Initially, Snuke is at the square (3,2). He can reach the square (3,4) by making five strokes as follows:

From (3,2), go west one square to (3,1).
From (3,1), go north two squares to (1,1).
From (1,1), go east two squares to (1,3).
From (1,3), go east one square to (1,4).
From (1,4), go south two squares to (3,4).

Samples2

Sample Input 2

1 6 4
1 1 1 6
......

Sample Output 2

Samples3

Sample Input 3

3 3 1
2 1 2 3
.@.
.@.
.@.

Sample Output 3

-1

Constraints

1 ≤ H, W, K ≤10^6
H×W ≤ 10^6
1 ≤ x1, x2 ≤ H
1 ≤ y1, y2 ≤ W
x1 ≠ x2 or y1≠y2.
ci,j is . or @.
cx1,y1= .
cx2,y2= .
All numbers in input are integers.

题解报告

题目翻译

Snuke，水上平衡车，住在一个矩形池塘，可以看成 H 列 W 行，(i, j) 表示第 i 列第 j 行。池塘里长着荷叶，荷叶是不能进入的。如果 cij 是 @，表示荷叶。如果 cij 是 .，表示不是荷叶。

Snuke 每次可以向北、东、南、西的任意同一个方向移动一步到 K 步，但是不能通过荷叶，同时也不能移动到池塘外。

给我们起点坐标 (x1, y1) 和终点坐标 (x2, y2)，要求我们找到最小的移动步数。如果不能到达，输出 -1。

好吧。写到这里，熟悉的 BFS 模板题。和标准的 BFS 不同的地方在于，标准 BFS 一次走一格，而本题一次可以走 1~K 以内的任意格。

样例数据分析

AC 参考代码

#include <bits/stdc++.h>
using namespace std;int main() {//读入数据int h, w, k;scanf("%d%d%d", &h, &w, &k);int x1, y1, x2, y2;scanf("%d%d%d%d", &x1, &y1, &x2, &y2);//迷宫描述char maze[h][w];//距离描述int dis[h][w];for (int i=0; i<h; i++) {for (int j=0; j<w; j++) {scanf(" %c", &maze[i][j]);dis[i][j] = INT_MAX;}}//数据处理//由于我们是从0开始x1--;y1--;x2--;y2--;const int dir[4][2] = {{-1,0}, {0,-1}, {1,0}, {0,1}};//方向定义次序无所谓queue<pair<int, int> > q;//插入起点dis[x1][y1] = 0;q.push({x1, y1});while (!q.empty()) {//获取队首数据int x = q.front().first;int y = q.front().second;q.pop();//按照规定方向移动for (int i=0; i<4; i++) {//新坐标和距离 int nx = x;int ny = y;int nd = dis[x][y]+1;//1 ~ k步for (int j=1; j<=k; j++) {//计算新的坐标nx += dir[i][0];ny += dir[i][1];//合法性判断if (nx<0 || nx>=h || ny<0 || ny>=w || '@'==maze[nx][ny] || dis[nx][ny]<nd) {break;}if (dis[nx][ny]>nd) {dis[nx][ny] = nd;q.push({nx, ny});}}}}int ans = dis[x2][y2];printf("%d\n", (INT_MAX==ans ? -1 : ans));return 0;
}

P.S.

难道是我碰到的最简单的 F 题。

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