hdu 2612 Find a way 【bfs】

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概述

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2612

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24234 Accepted Submission(s): 7933

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

题意：两个人分别从两个位置出发，求到达终点的时间加和！（终点可能不止一个）

思路：两次bfs+队列的使用，开两个时间数组t1和t2,分别记录两个人各自经过目的点的时间，最后判断一下！

#include<iostream>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;
#define ms(a) memset(a,0,sizeof a)
const int inf = 0x3f3f3f3f;
#define ma(b) memset(b,inf,sizeof b)
const int maxn = 210;
char map1[maxn][maxn];
int vis[maxn][maxn],n,m,x2,y2,x0,y0;
int t1[maxn][maxn],t2[maxn][maxn];
struct node{
int x,y,step;
};
int dis[4][2]={0,1,0,-1,1,0,-1,0};
bool check(int x,int y)
{
if(x>=0 && x<n && y>=0 && y<m)
return true;
return false;
}
void bfs(int x1,int y1,int q1)
{
queue<node>q;
node e1,e2;
e1.x = x1;
e1.y = y1;
e1.step = 0;
vis[e1.x][e1.y] = 1;
q.push(e1);
while(!q.empty())
{
e2 = q.front();
q.pop();
if(map1[e2.x][e2.y] == '@')
{
if(q1==1)
t1[e2.x][e2.y] = e2.step;
else
t2[e2.x][e2.y] = e2.step;
}
for(int i=0;i<4;i++)
{
e1.x = e2.x + dis[i][0];
e1.y = e2.y + dis[i][1];
if(check(e1.x,e1.y) && map1[e1.x][e1.y]!='#' && !vis[e1.x][e1.y])
{
e1.step = e2.step + 1;
vis[e1.x][e1.y] = 1;
q.push(e1);
}
}
}
}
int main()
{
while(cin>>n>>m)
{
for(int i=0;i<n;i++)
cin>>map1[i];
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(map1[i][j] == 'Y')
{
x0 = i;
y0 = j;
}
if(map1[i][j] == 'M')
{
x2 = i;
y2 = j;
}
}
}
ms(vis);
ma(t1);
bfs(x0,y0,1);
ms(vis);
ma(t2);
bfs(x2,y2,2);
int min1 = inf;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(t1[i][j]!=inf && t2[i][j]!=inf)
{
min1 = min(t1[i][j]+t2[i][j],min1);
}
}
}
cout<<min1*11<<endl;
}
return 0;
}