HDU——2612Find a way（多起点多终点BFS）

最新推荐文章于 2022-01-21 13:00:34 发布

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最新推荐文章于 2022-01-21 13:00:34 发布

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文章标签： java

原文链接：http://www.cnblogs.com/Blackops/p/5766306.html

本文介绍了一个名为 Findaway 的最短路径算法问题，该问题要求找出两个角色从初始位置到同一目标位置（KFC）的最短总时间。通过使用广度优先搜索策略，算法有效地计算了所有可能路径中的最小总时间。

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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\n",r);
	}
	return 0;
}

转载于:https://www.cnblogs.com/Blackops/p/5766306.html