[swustoj 1023] Escape

Escape

 

 

Description

BH is in a maze,the maze is a matrix,he wants to escape!

Input

The input consists of multiple test cases.

For each case,the first line contains 2 integers N,M( 1 <= N, M <= 100 ).

Each of the following N lines contain M characters. Each character means a cell of the map.

Here is the definition for chracter.

For a character in the map: 

'S':BH's start place,only one in the map.

'E':the goal cell,only one in the map.

'.':empty cell.

'#':obstacle cell.

'A':accelerated rune.

BH can move to 4 directions(up,down,left,right) in each step.It cost 2 seconds without accelerated rune.When he get accelerated rune,moving one step only cost 1 second.The buff lasts 5 seconds,and the time doesn't stack when you get another accelerated rune.(that means in anytime BH gets an accelerated rune,the buff time become 5 seconds).

Output

The minimum time BH get to the goal cell,if he can't,print "Please help BH!".

Sample Output

5 5

....E

.....

.....

##...

S#...

5 8

........

........

..A....A

A######.

S......E

Sample Output
Please help BH!
12

BFS、注意一下优先级判断就行了

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;
#define N 110

struct Node
{
    int x,y,t,r; //坐标,时间,剩余加速时间
    bool operator <(const Node &T)const{
        if(t!=T.t)return t>T.t;
        return r<T.r;
    }
};

int n,m;
int sx,sy;
int vis[N][N][8];
char mpt[N][N];
int dir[4][2]={-1,0,0,1,0,-1,1,0};

void bfs()
{
    Node now,next;
    priority_queue<Node> q;
    memset(vis,0,sizeof(vis));
    now.x=sx;
    now.y=sy;
    now.t=now.r=0;
    vis[sx][sy][0]=1;
    q.push(now);
    while(!q.empty())
    {
        now=q.top();
        q.pop();
        if(mpt[now.x][now.y]=='E'){
            printf("%d
",now.t);
            return;
        }
        for(int i=0;i<4;i++){
            next=now;
            next.x+=dir[i][0];
            next.y+=dir[i][1];
            next.t+=2;
            if(next.r>=1) {next.t--;next.r--;}
            if(next.x>=1 && next.x<=n && next.y>=1 && next.y<=m && mpt[next.x][next.y]!='#'){
                if(mpt[next.x][next.y]=='A') next.r=5;
                if(!vis[next.x][next.y][next.r]){
                    vis[next.x][next.y][next.r]=1;
                    q.push(next);
                }
            }
        }
    }
    printf("Please help BH!
");
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf(" %c",&mpt[i][j]);
                if(mpt[i][j]=='S'){
                    sx=i;
                    sy=j;
                }
            }
        }
        bfs();
    }
    return 0;
}