1685: Route Planning

题目描述

然而事情并不是Richard想的那么简单，等他登陆后他才发现，地图里有各种各样的炸弹和关卡。当Richard经过炸弹的时候，他需要花费1分钟的时间拆弹；地图中有关卡对应的通行证，如果Richard拿到了通行证，那么它就可以通过对应的关卡。现在请你重新为Richard规划一条路线，使执行任务的时间最短。

.代表空地

‘#’代表障碍

@代表出发点

*代表任务地点

=代表炸弹

1-4代表通行证

a-d代表关卡

输入

输入第一行为T，即有T组输入数据（1<=T<=10）

每组数据第一行为整数N 10≤N≤100，代表地图的长度和宽度

接下来为N行，每行N个字符，代表地图

输出

输出一个整数，为Richard从起点出发到达终点所需的最短时间。

如果Richard无法到达终点，请输出MISSION FAILED

每组输出占一行

样例输入

5
@….
.1…
…..
=a###
….*

样例输出

8

---

传入节点不会段溢出

---

#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;}