zoukankan      html  css  js  c++  java
  • HDU——2612Find a way(多起点多终点BFS)

    Find a way

    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 9698    Accepted Submission(s): 3165

    Problem Description
    Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki.
    Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest. 
    Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.
     
    Input
    The input contains multiple test cases.
    Each test case include, first two integers n, m. (2<=n,m<=200). 
    Next n lines, each line included m character.
    ‘Y’ express yifenfei initial position.
    ‘M’    express Merceki initial position.
    ‘#’ forbid road;
    ‘.’ Road.
    ‘@’ KCF
     
    Output
    For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.
     
    Sample Input
    4 4 Y.#@ .... .#.. @..M 4 4 Y.#@ .... .#.. @#.M 5 5 Y..@. .#... .#... @..M. #...#
     
    Sample Output
    66 88 66
     

    主要看起点和终点那种点少,以少的为起点开搜。1A水过

    #include<iostream>
    #include<algorithm>
    #include<cstdlib>
    #include<sstream>
    #include<cstring>
    #include<cstdio>
    #include<string>
    #include<deque>
    #include<stack>
    #include<cmath>
    #include<queue>
    #include<set>
    #include<map>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define MM(x) memset(x,0,sizeof(x))
    #define MMINF(x) memset(x,INF,sizeof(x))
    typedef long long LL;
    const double PI=acos(-1.0);
    const int N=210;
    char pos[N][N];
    int vis[N][N];
    int n,m;
    struct info
    {
    	int x;
    	int y;
    	int time;
    	bool operator<(const info &a)
    	{
    		return time>a.time;
    	}
    };
    vector<info>kfc;
    int total[2][200][200];
    info direct[4]={{1,0,0},{-1,0,0},{0,1,0},{0,-1,0}},S,T;
    inline info operator+(const info &a,const info &b)
    {
    	info c;
    	c.x=a.x+b.x;
    	c.y=a.y+b.y;
    	return c;
    }
    void init()
    {
    	MM(pos);
    	MM(vis);
    	MMINF(total);
    	kfc.clear();
    }
    inline bool check(const info &a)
    {
    	return (a.x>=0&&a.x<n&&a.y>=0&&a.y<m&&!vis[a.x][a.y]&&pos[a.x][a.y]!='#');
    }
    queue<info>Q;
    int main(void)
    {
    	int i,j,cnt;
    	while (~scanf("%d%d",&n,&m))
    	{
    		init();
    		cnt=0;
    		for (i=0; i<n; i++)
    		{
    			for (j=0; j<m; j++)
    			{
    				cin>>pos[i][j];
    				if(pos[i][j]=='Y')
    				{
    					S.x=i;
    					S.y=j;
    					S.time=0;
    				}
    				else if(pos[i][j]=='M')
    				{
    					T.x=i;
    					T.y=j;
    					T.time=0;
    				}
    				else if(pos[i][j]=='@')
    				{
    					info k;
    					k.x=i;
    					k.y=j;
    					kfc.push_back(k);
    					cnt++;
    				}
    			}
    		}
    		int c=0;
    		while (!Q.empty())
    			Q.pop();
    		Q.push(S);
    		vis[S.x][S.y]=1;
    		while (!Q.empty())
    		{
    			info now=Q.front();
    			Q.pop();
    			if(pos[now.x][now.y]=='@')
    			{
    				total[0][now.x][now.y]=now.time;
    				if(++c==cnt)
    					break;
    			}	
    			for (i=0; i<4; i++)
    			{
    				info v=now+direct[i];
    				if(check(v))
    				{
    					v.time=now.time+11;
    					vis[v.x][v.y]=1;
    					Q.push(v);
    				}				
    			}
    		}
    		while (!Q.empty())
    			Q.pop();
    		MM(vis);
    		c=0;
    		Q.push(T);
    		while (!Q.empty())
    		{
    			info now=Q.front();
    			Q.pop();
    			if(pos[now.x][now.y]=='@')
    			{
    				total[1][now.x][now.y]=now.time;
    				if(++c==cnt)
    					break;
    			}	
    			for (i=0; i<4; i++)
    			{
    				info v=now+direct[i];
    				if(check(v))
    				{
    					v.time=now.time+11;
    					vis[v.x][v.y]=1;
    					Q.push(v);
    				}				
    			}
    		}
    		int r=INF;
    		for (i=0; i<cnt; i++)
    			r=min(r,total[0][kfc[i].x][kfc[i].y]+total[1][kfc[i].x][kfc[i].y]);
    		printf("%d
    ",r);
    	}
    	return 0;
    }
  • 相关阅读:
    Git原理与命令大全
    【网络安全】加解密算法最详解
    陪你阅读《区块链:从数字货币到信用社会》序一
    Splunk初识
    红帽学习记录[RHCE] ISCSI远程块储存
    DNS 域名系统与邮件服务器
    红帽学习记录[RHCE] 防火墙与网络合作
    红帽学习笔记[RHCE]网络配置与路由转发
    红帽学习笔记[RHCE]OpenLDAP 服务端与客户端配置
    红帽学习笔记[RHCSA] 第二周
  • 原文地址:https://www.cnblogs.com/Blackops/p/5766306.html
Copyright © 2011-2022 走看看