zoukankan      html  css  js  c++  java
  • hdu2579之BFS

    Dating with girls(2)

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1470    Accepted Submission(s): 414

    Problem Description
    If you have solved the problem Dating with girls(1).I think you can solve this problem too.This problem is also about dating with girls. Now you are in a maze and the girl you want to date with is also in the maze.If you can find the girl, then you can date with the girl.Else the girl will date with other boys. What a pity! 
    The Maze is very strange. There are many stones in the maze. The stone will disappear at time t if t is a multiple of k(2<= k <= 10), on the other time , stones will be still there. 
    There are only ‘.’ or ‘#’, ’Y’, ’G’ on the map of the maze. ’.’ indicates the blank which you can move on, ‘#’ indicates stones. ’Y’ indicates the your location. ‘G’ indicates the girl's location . There is only one ‘Y’ and one ‘G’. Every seconds you can move left, right, up or down.
     
    Input
    The first line contain an integer T. Then T cases followed. Each case begins with three integers r and c (1 <= r , c <= 100), and k(2 <=k <= 10).
    The next r line is the map’s description.
     
    Output
    For each cases, if you can find the girl, output the least time in seconds, else output "Please give me another chance!".
     
    Sample Input
    1 6 6 2 ...Y.. ...#.. .#.... ...#.. ...#.. ..#G#.
     
    Sample Output
    7
    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<string>
    #include<queue>
    #include<algorithm>
    #include<map>
    #include<iomanip>
    #define INF 99999999
    using namespace std;
    
    const int MAX=100+10;
    char Map[MAX][MAX];
    int mark[MAX][MAX][20];
    int n,m,k;
    int dir[4][2]={0,1,0,-1,1,0,-1,0};
    
    struct Node{
    	int x,y,time;
    	Node(){}
    	Node(int X,int Y,int Time):x(X),y(Y),time(Time){}
    }start;
    
    int BFS(int &flag){
    	queue<Node>q;
    	Node oq,next;
    	q.push(start);
    	mark[start.x][start.y][start.time%k]=flag;
    	while(!q.empty()){
    		oq=q.front();
    		q.pop();
    		for(int i=0;i<4;++i){
    			next=Node(oq.x+dir[i][0],oq.y+dir[i][1],oq.time+1);
    			if(next.x<0 || next.y<0 || next.x>=n || next.y>=m)continue;
    			if(mark[next.x][next.y][next.time%k] == flag)continue;
    			if(next.time%k != 0 && Map[next.x][next.y] == '#')continue;
    			mark[next.x][next.y][next.time%k]=flag;
    			if(Map[next.x][next.y] == 'G')return next.time;
    			q.push(next);
    		}
    	}
    	return -1;
    }
    
    int main(){
    	int num=0,t;
    	cin>>t;
    	while(t--){
    		cin>>n>>m>>k;
    		for(int i=0;i<n;++i)cin>>Map[i];
    		for(int i=0;i<n;++i){
    			for(int j=0;j<m;++j){
    				if(Map[i][j] == 'Y')start.x=i,start.y=j;
    			}
    		}
    		start.time=0;
    		int temp=BFS(++num);
    		if(temp != -1)cout<<temp<<endl;
    		else cout<<"Please give me another chance!"<<endl;
    	}
    	return 0;
    }



  • 相关阅读:
    c++读写MySQL
    感叹游戏行业的飞速发展
    和真正的程序员在一起是怎样的体验
    程序媛是怎样找老公的
    IO和socket编程
    郁金香搜索引擎的方案
    实现一个自己的搜索引擎的初始规划
    JVM知识在离线数据中的运用
    看Lucene源码必须知道的基本规则和算法
    看Lucene源码必须知道的基本概念
  • 原文地址:https://www.cnblogs.com/riskyer/p/3221825.html
Copyright © 2011-2022 走看看