hdu 2612 Find a way（BFS + 最短路）

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

Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 10934    Accepted Submission(s): 3585

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

 

Author

yifenfei

 

Source

奋斗的年代

 

Recommend

yifenfei

 

Statistic | Submit | Discuss | Note

题目大意： yifenfei 和 Merceki 想要见面，找一个最近的 KCF ，求他们总共需要走的最近距离，Y代表  yifenfei
的家， M 代表 Merceki 的家，

 
             @代表KFC，#代表墙，.代表能走的路

解析：开两个数组分别存这两个人到任意地点要走的路程，两边BFS，然后就下到@的最小值，输出即可

代码如下：

#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<string>
#include<cstdio>
#include<cstring>
#include<cctype>
#include<cmath>
#define N 209
using namespace std;
const int inf = 1e9;
const int mod
c261
= 1<<30;
const double eps = 1e-8;
const double pi = acos(-1.0);
typedef long long LL;
char mp[210][210];
int m, n, a

, b

;
int nx[11] = {-1, 0, 0, 1};
int ny[11] = {0, -1, 1, 0};
struct node
{
int x, y;
};
void bfs(int s, int e)
{
int i, x, y;
a[s][e] = 0;
queue<node> q;
node now, nn;
now.x = s; now.y = e;
q.push(now);
while(!q.empty())
{
now = q.front(); q.pop();
for(i = 0; i < 4; i++)
{
x = now.x + nx[i]; y = now.y + ny[i];
if(x < 1 || y < 1 || x > m || y > n) continue;
if(a[x][y] != -1 || mp[x][y] == '#') continue;
a[x][y] = a[now.x][now.y] + 1;
nn.x = x; nn.y = y;
q.push(nn);
}
}
}

void Bfs(int s, int e)
{
int i, x, y;
b[s][e] = 0;
queue<node> q;
node now, nn;
now.x = s; now.y = e;
q.push(now);
while(!q.empty())
{
now = q.front(); q.pop();

for(i = 0; i < 4; i++)
{
x = now.x + nx[i]; y = now.y + ny[i];
if(x < 1 || y < 1 || x > m || y > n) continue;
if(b[x][y] != -1 || mp[x][y] == '#') continue;
b[x][y] = b[now.x][now.y] + 1;
nn.x = x; nn.y = y;
q.push(nn);
}
}
}

int main()
{
int i, j, Y1, Y2, M1, M2;
while(~scanf("%d%d", &m, &n))
{
for(i = 1; i <= m; i++)
{
for(j = 1; j <= n; j++)
{
scanf(" %c", &mp[i][j]);
if(mp[i][j] == 'Y') Y1 = i, Y2 = j;
else if(mp[i][j] == 'M') M1 = i, M2 = j;
}
}
memset(a, -1 ,sizeof(a));
memset(b, -1, sizeof(b));
bfs(Y1, Y2);
Bfs(M1, M2);
int ans = inf;
for(i = 1; i <= m; i++)
{
for(j = 1; j <= n; j++)
{
if(mp[i][j] == '@' && a[i][j] != -1 && b[i][j] != -1) ans = min(ans, a[i][j] + b[i][j]);//-1表示不能到达该KFC
}
}
printf("%d\n", ans * 11);
}
return 0;
}