zoukankan      html  css  js  c++  java
  • 1242Rescue (优先队列BFS)

    Problem Description
    Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.

    Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.

    You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)
     
    Input
    First line contains two integers stand for N and M.

    Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.

    Process to the end of the file.
     
    Output
    For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life."
     
    Sample Input
    7 8 #.#####. #.a#..r. #..#x... ..#..#.# #...##.. .#...... ........
     
    Sample Output
    13
     
    #include<stdio.h>
    #include<iostream>
    #include<queue>
    using namespace std;
    typedef struct nn
    {
         int x,y;
        int time;
        friend bool operator<(nn n1,nn n2)
        {
            return n2.time<n1.time;
        }
    }node;
    int h,w,si,sj,di,dj,minT,flog;
    int dir[4][2]={{1,0},{-1,0},{0,-1},{0,1}};
    char map[205][205];
    void BFS(void)
    {
        int tx,ty,e,k,i;
        priority_queue<node> Q;
        node q,p;
    
        q.time=0;q.x=sj;q.y=si;
        Q.push(q);
        while(!Q.empty())
        {
            q=Q.top();
            Q.pop();
            for(e=0;e<4;e++)
            {
                tx=q.x+dir[e][1];
                ty=q.y+dir[e][0];
                if(tx>=0&&tx<w&&ty>=0&&ty<h&&map[ty][tx]!='#')
                {
                    p.time=q.time+1; p.x=tx; p.y=ty;
                    if(p.y==di&&p.x==dj)
                    {
                       minT=p.time;
                        flog=1;
                        return ;
                    }
                    if(map[ty][tx]=='x')
                    {
                        p.time+=1;//printf("#");
                    }
                    map[ty][tx]='#';
                    Q.push(p);
                }
            }
        }
    }
    int main()
    {
        int i,j;
        while(scanf("%d%d",&h,&w)>0)
        {
        for(i=0;i<h;i++)
        {
            getchar();
            for(j=0;j<w;j++)
            {
                scanf("%c",&map[i][j]);
                if(map[i][j]=='r')
                {
                    si=i;sj=j;
                }
                if(map[i][j]=='a')
                {
                    di=i;dj=j;
                }
            }
        }
        flog=0;
        BFS();
        if(flog==0)
        printf("Poor ANGEL has to stay in the prison all his life.
    ");
        else
        printf("%d
    ",minT);
        }
    }
    



  • 相关阅读:
    python操作json来存储简单的数据,pickle来操作复杂的数据
    python元组,列表,字典练习
    python文件实现增删改查操作
    socket模拟服务器,客户端下载东西(ftp)
    optiontransferselect标签
    doubleselect标签联动选择框
    datetimepicker标签日期选择器
    combobox标签复合框
    checkboxlist标签多选框组
    使用动态数据的autocomplete标签
  • 原文地址:https://www.cnblogs.com/riskyer/p/3233776.html
Copyright © 2011-2022 走看看