bsf find a way
Description
Input
Output
Sample Input
Sample Output
Hint
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.
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
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
感谢汤神 我的一直 wa 汤神不辞幸苦帮我写了一个 真的是 万分感谢
#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<queue>
#define MAX 0x3f3f3f3f
using namespace std;
char map[205][205];
struct node
{
int x;
int y;
int step;
};
int n, m;
int dis[205][205];
int nextd[4][2] = { 0,1,0,-1,1,0,-1,0 };
void bfs(int x, int y)
{
bool vis[205][205];
memset(vis, false, sizeof(vis));
queue<node>q;
node t, s;
vis[x][y] = true;
s.x = x;
s.y = y;
s.step = 0;
q.push(s);
while (!q.empty())
{
s = q.front();
q.pop();
if (map[s.x][s.y] == '@')
{
if (dis[s.x][s.y] == MAX)
dis[s.x][s.y] = s.step;
else
dis[s.x][s.y] += s.step;
}
for (int i = 0; i < 4; i++)
{
t.x = s.x + nextd[i][0];
t.y = s.y + nextd[i][1];
t.step = s.step + 1;
if (t.x < 0 || t.x >= n || t.y < 0 || t.y >= m) continue;
if (vis[t.x][t.y]) continue;
if (map[t.x][t.y] == '#') continue;
vis[t.x][t.y] = true;
q.push(t);
}
}
}
int main()
{
node M, Y;
int i, j;
while (~scanf("%d %d", &n, &m))
{
memset(dis, 0x3f, sizeof(dis));
for (i = 0; i < n; i++) scanf("%s", map[i]);
for (i = 0; i < n; i++)
{
for (j = 0; j < m; j++)
{
if (map[i][j] == 'M')
{
M.x = i;
M.y = j;
}
if (map[i][j] == 'Y')
{
Y.x = i;
Y.y = j;
}
}
}
bfs(M.x, M.y);
bfs(Y.x, Y.y);
int min = MAX;
for (i = 0; i < n; i++)
{
for (j = 0; j < m; j++)
{
if (min <= dis[i][j]) continue;
min = dis[i][j];
}
}
printf("%d\n", min*11);
}
return 0;
}
by swust.acm tzc
查看6道真题和解析
