Find a way
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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
奮鬥的年代
這幾天忙學習... 找了些題保持手感。
思路：2次BFS 得到Y、M 到@的最短距離
ps ：測了下杭電沒給Y、M不能互走，或者互走影響結果的數據，注意一點，可能KFC周圍都是牆。
有幾個細節可以優化，我這沒加工。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define min(a,b) (a>b?b:a)
#define CLR(arr,x) memset(arr,x,sizeof(arr))
using namespace std;
const int maxn = 400 ;
const int inf = 1<<30 ;
char mapp[maxn][maxn] ;
int ans1[maxn][maxn] , ans2[maxn][maxn] ;
bool step[maxn][maxn] ;
int Next[4][2] = { {0,1} , {1,0} , {0,-1} ,{-1,0} } ;
int n , m ;
struct point{
int row , col ;
};
point st , st2 ;
void init() {
CLR(ans1,0) ;
CLR(ans2,0) ;
CLR(mapp,0) ;
}
bool judge(point init) {
if( init.row>=0&&init.row<n && init.col>=0 && init.col < m ) {
return true ;
}
return false;
}
void bfs(point temp) {
queue<point> V ;
while(!V.empty()){
V.pop() ;
}
V.push(temp) ;
CLR(step,0) ;
step[temp.row][temp.col] = true ;
point init ;
while(!V.empty()) {
temp = V.front() ;
V.pop() ;
for( int i = 0 ; i < 4 ; ++i) {
init.row = temp.row+Next[i][0] ;
init.col = temp.col+Next[i][1] ;
if( judge(init) && !step[init.row][init.col] ){
if( mapp[init.row][init.col] != '#' ) {
ans1[init.row][init.col] = ans1[temp.row][temp.col]+1 ;
step[init.row][init.col] = true ;
V.push(init) ;
}
}
}
}
}
void bfs2(point temp) {
queue<point> V2 ;
while(!V2.empty()){
V2.pop() ;
}
V2.push(temp) ;
CLR(step,0) ;
step[temp.row][temp.col] = true ;
point init ;
while(!V2.empty()) {
temp = V2.front() ;
V2.pop() ;
for( int i = 0 ; i < 4 ; ++i) {
init.row = temp.row+Next[i][0] ;
init.col = temp.col+Next[i][1] ;
if( judge(init) && !step[init.row][init.col] ){
if( mapp[init.row][init.col] != '#' ) {
ans2[init.row][init.col] = ans2[temp.row][temp.col]+1 ;
V2.push(init) ;
step[init.row][init.col] = true ;
}
}
}
}
}
int main() {
while(~scanf("%d%d",&n,&m)) {
init() ;
for( int i = 0 ; i < n ; ++i ){
scanf("%s",mapp[i]) ;
for( int j = 0 ; j < m ; ++j ){
if( mapp[i][j] == 'Y') {
st.row = i ;
st.col = j ;
}
if(mapp[i][j] == 'M') {
st2.row = i ;
st2.col = j ;
}
}
}
bfs(st) ;
bfs2(st2) ;
int minn = inf ;
for( int i = 0 ;i < n ; ++i ) {
for( int j = 0 ;j < m ; ++j ){
if(mapp[i][j] == '@' && ans1[i][j] && ans2[i][j] ) {
minn = min(minn,ans1[i][j]+ans2[i][j]) ;
}
}
}
printf("%d\n",minn*11) ;
}
return 0 ;
}