1685: Route Planning

題目描述

然而事情並不是Richard想的那麼簡單，等他登陸後他才發現，地圖裏有各種各樣的炸彈和關卡。當Richard經過炸彈的時候，他需要花費1分鐘的時間拆彈；地圖中有關卡對應的通行證，如果Richard拿到了通行證，那麼它就可以通過對應的關卡。現在請你重新爲Richard規劃一條路線，使執行任務的時間最短。

.代表空地

‘#’代表障礙

@代表出發點

*代表任務地點

=代表炸彈

1-4代表通行證

a-d代表關卡

輸入

輸入第一行爲T，即有T組輸入數據（1<=T<=10）

每組數據第一行爲整數N 10≤N≤100，代表地圖的長度和寬度

接下來爲N行，每行N個字符，代表地圖

輸出

輸出一個整數，爲Richard從起點出發到達終點所需的最短時間。

如果Richard無法到達終點，請輸出MISSION FAILED

每組輸出佔一行

樣例輸入

5
@….
.1…
…..
=a###
….*

樣例輸出

8

---

傳入節點不會段溢出

---

#include<iostream>
#include<queue>
#include<cstring>
#include <cstdio>
#include<memory.h>

using namespace std;

const int N=120;
int ans, n, treax[10], treay[10], treal;
bool a[N][N][20][64];
int fx[] = {0, 1, 0, -1, 0}, fy[] = {0, 0, 1, 0, -1};
char maze[N][N];

struct node
{
    int x, y, time;
    int board, trea;
};

int judgenum(int x, int y)
{
    for(int i = 1; i <= treal; i++)
    {
        if(treax[i] == x && treay[i] == y )
        {
            return i;
        }
    }
}

node New, now, first;

void bfs(node xxx)
{
    queue<node> q;


    a[first.x][first.y][first.trea][first.board] = true;

    ans = 9999;

    q.push(first);
    while(!q.empty())
    {
        now = q.front();
        q.pop();


        if(maze[now.x][now.y] == '*')
        {
            ans = min(now.time, ans);
            continue;
        }
        for(int i = 1; i <= 4; i++)
        {

            New = now;
            New.x += fx[i];
            New.y += fy[i];
            New.time ++ ;

            if(New.x < 1 || New.y < 1 || New.x > n || New.y > n || maze[New.x][New.y] == '#')
                continue;

            char ch = maze[New.x][New.y];

            if(ch == '=')
            {
                int xx = judgenum(New.x, New.y);

                if((New.trea & (1<<xx)) == 0)
                    New.trea |= (1<<xx);

                if(!a[New.x][New.y][New.trea][New.board])
                {
                    a[New.x][New.y][New.trea][New.board] = true;
                    New.time ++;
                    q.push(New);
                }
            }

            else if(ch >= '1' && ch <= '4')
            {
                int j = maze[New.x][New.y] - '1';
                    New.board |= (1<<j);

                if(!a[New.x][New.y][New.trea][New.board])
                {
                    q.push(New);
                    a[New.x][New.y][New.trea][New.board] = true;
                }
            }

            else if(ch >= 'a' && ch <= 'd')
            {
                if(New.board & (1<<(ch-'a')) && !a[New.x][New.y][New.trea][New.board])
                {
                    a[New.x][New.y][New.trea][New.board] = true;
                    q.push(New);
                }
            }

            else if(!a[New.x][New.y][New.trea][New.board])
            {
                a[New.x][New.y][New.trea][New.board] = true;
                q.push(New);
            }
        }
    }
}

int main()
{
    int t, x, y;

        while(scanf("%d", &n) != EOF){
            //scanf("%d", &n);
            memset(a, 0, sizeof(a));
            treal = 0;

            for(int i = 1; i <= n; i++)
            {
                scanf("%s", &maze[i][1]);
                for(int j = 1; j <= n; j++)
                {
                    if(maze[i][j] == '@')
                    {
                        first.x = i;
                        first.y = j;
                    }
                    else if(maze[i][j] == '=')
                    {
                        treax[++treal] = i; treay[treal] = j;
                    }
                }
            }

            first.time = 0;
            first.board = 0;
            first.trea = 0;

            bfs(first);
            if(ans == 9999)
                printf("MISSION FAILED\n");
            else
                printf("%d\n", ans);
        }

    //}
    return 0;
}