AtCoder Beginner Contest 176 D-Wizard in Maze(双端队列)

Sample Input 1
Copy

4 4
1 1
4 4
..#.
..#.
.#..
.#..

Sample Output 1

Note that he cannot walk diagonally.

Sample Input 2
Copy

4 4
1 4
4 1
.##.
####
####
.##.

Sample Output 2
Copy
-1
He cannot move from there.

Sample Input 3

4 4
2 2
3 3
…
…
…
…

Sample Output 3

0

No use of magic is needed.

Sample Input 4

4 5 1 2 2 5
#.###
####.
#…##
#…##

Sample Output 4

2

分析：
对于迷宫类问题显然要用BFS解决。因为要求使用魔法次数最少，所以要尽可能先用A方式移动。

因此BFS时可以采用双端队列，将不使用魔法的状态放队首、使用魔法的状态放队尾。（其实也能用优先队列）

想到这个代码就很好写了。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<time.h>
#include<vector>
using namespace std;
struct node
{
    int x,y;
    int step;
};
int h,w,sx,sy,ex,ey;
deque<node> q;
char s[1005][1005],vis[1005][1005];
int nextx[]={0,-1,0,1},nexty[] = {-1,0,1,0};
int main()
{
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    scanf("%d%d",&h,&w);
    scanf("%d%d",&sx,&sy);
    scanf("%d%d",&ex,&ey);
    for(int i=1; i<=h; ++i){
        for(int j=1; j<=w; ++j)
            scanf(" %c",&s[i][j]);
    }
    /*
    for(int i=1; i<=h; ++i){
        for(int j=1; j<=w; ++j)
            printf("%c",s[i][j]);
        putchar('\n');
    }*/

    vis[sx][sy] = 1;
    q.push_back((node){sx,sy,0});
    //cout<<q.front().x<<' '<<q.front().y;
    //getchar();
    int ans = -1;
    while(!q.empty()){
        struct node t = q.front();
        q.pop_front();
        vis[t.x][t.y] = 1;
        if(t.x==ex&&t.y==ey){
            ans = t.step;
            break;
        }

        for(int i=0; i<4; ++i){
            int tx,ty;
            tx = t.x + nextx[i];
            ty = t.y + nexty[i];
            if(tx>0&&tx<=h&&ty>0&&ty<=w&&s[tx][ty]=='.'&&!vis[tx][ty]){
                q.push_front((node){tx,ty,t.step});
            }
        }

        for(int tx=t.x-2; tx<=t.x+2; ++tx){
            for(int ty=t.y-2; ty<=t.y+2; ++ty){
                if(tx>0&&tx<=h&&ty>0&&ty<=w&&s[tx][ty]=='.'&&!vis[tx][ty]){
                    if(tx==t.x&&ty==t.y)
                        continue;
                    q.push_back((node){tx,ty,t.step+1});
                }
            }
        }
    }
    printf("%d\n",ans);
    return 0;
}

本文地址：https://blog.csdn.net/qq_26139541/article/details/108558933