几个比较好的搜索题
HDU 3309 Roll The Cube
Problem Description
This is a simple game.The goal of the game is to roll two balls to two holes each.
'B' -- ball
'H' -- hole
'.' -- land
'*' -- wall
Remember when a ball rolls into a hole, they(the ball and the hole) disappeared, that is , 'H' + 'B' = '.'.
Now you are controlling two balls at the same time.Up, down , left , right --- once one of these keys is pressed, balls exist roll to that direction, for example , you pressed up , two balls both roll up.
A ball will stay where it is if its next point is a wall, and balls can't be overlap.
Your code should give the minimun times you press the keys to achieve the goal.
'B' -- ball
'H' -- hole
'.' -- land
'*' -- wall
Remember when a ball rolls into a hole, they(the ball and the hole) disappeared, that is , 'H' + 'B' = '.'.
Now you are controlling two balls at the same time.Up, down , left , right --- once one of these keys is pressed, balls exist roll to that direction, for example , you pressed up , two balls both roll up.
A ball will stay where it is if its next point is a wall, and balls can't be overlap.
Your code should give the minimun times you press the keys to achieve the goal.
Input
First there's an integer T(T<=100) indicating the case number.
Then T blocks , each block has two integers n , m (n , m <= 22) indicating size of the map.
Then n lines each with m characters.
There'll always be two balls(B) and two holes(H) in a map.
The boundary of the map is always walls(*).
Then T blocks , each block has two integers n , m (n , m <= 22) indicating size of the map.
Then n lines each with m characters.
There'll always be two balls(B) and two holes(H) in a map.
The boundary of the map is always walls(*).
Output
The minimum times you press to achieve the goal.
Tell me "Sorry , sir , my poor program fails to get an answer." if you can never achieve the goal.
Tell me "Sorry , sir , my poor program fails to get an answer." if you can never achieve the goal.
Sample Input
4 6 3 *** *B* *B* *H* *H* *** 4 4 **** *BB* *HH* **** 4 4 **** *BH* *HB* **** 5 6 ****** *.BB** *.H*H* *..*.* ******
Sample Output
3 1 2 Sorry , sir , my poor program fails to get an answer.
const int Max_N = 25 ;
const int inf = (1<<30) ;
char table[Max_N][Max_N] ;
int minstep[Max_N][Max_N][Max_N][Max_N] ;
int N , M ;
struct Point{
int x ;
int y ;
int IsOK ;
Point(){}
Point(int i , int j , int s):x(i) ,y(j) , IsOK(s){}
};
struct Node{
Point A ;
Point B ;
int step ;
Node(){}
Node(Point a , Point b , int s):A(a) ,B(b),step(s){}
};
int d[4][2] = {{1,0} , {0,1} , {-1,0} ,{0 ,-1}} ;
int cango(int x , int y){
return 1 <= x && x <= N && 1 <= y && y <= M ;
}
int bfs(Node s){
int i , j , x , y ;
for(i = 1 ; i <= N ; i++)
for(j = 1 ; j <= M ; j++)
for(x = 1 ; x <= N ; x++)
for(y = 1 ; y <= M ; y++)
minstep[i][j][x][y] = inf ;
minstep[s.A.x][s.A.y][s.B.x][s.B.y] = 0 ;
queue<Node> que ;
Node now ;
Point A , B , nA , nB ;
int ax , ay , bx , by ;
que.push(s) ;
while(! que.empty()){
now = que.front() ;
que.pop() ;
A = now.A ;
B = now.B ;
if(A.IsOK && B.IsOK)
return now.step ;
for(i = 0 ; i < 4 ; i++){
ax = A.x + d[i][0] ;
ay = A.y + d[i][1] ;
bx = B.x + d[i][0] ;
by = B.y + d[i][1] ;
if(table[ax][ay] == '*' && table[bx][by] == '*')
continue ;
if(table[ax][ay] == '*'){
ax = A.x ;
ay = A.y ;
}
if(table[bx][by] == '*'){
bx = B.x ;
by = B.y ;
}
if(!A.IsOK && !B.IsOK && ax == bx && ay == by)
continue ;
if(A.IsOK){
nA = A ;
nB.x = bx ;
nB.y = by ;
if(table[bx][by] == 'H' && !(A.x==bx&&A.y==by))
nB.IsOK = 1 ;
else
nB.IsOK = 0 ;
if(minstep[nA.x][nA.y][nB.x][nB.y] > now.step + 1){
que.push(Node(nA , nB , now.step+1)) ;
minstep[nA.x][nA.y][nB.x][nB.y] = now.step + 1 ;
}
}
else if(B.IsOK){
nB = B ;
nA.x = ax ;
nA.y = ay ;
if(table[ax][ay] == 'H' && !(B.x==ax&&B.y==ay))
nA.IsOK = 1 ;
else
nA.IsOK = 0 ;
if(minstep[nA.x][nA.y][nB.x][nB.y] > now.step+1){
que.push(Node(nA , nB , now.step+1)) ;
minstep[nA.x][nA.y][nB.x][nB.y] = now.step + 1 ;
}
}
else{
nA.x = ax ;
nA.y = ay ;
if(table[ax][ay] == 'H')
nA.IsOK = 1 ;
else
nA.IsOK = 0 ;
nB.x = bx ;
nB.y = by ;
if(table[bx][by] == 'H')
nB.IsOK = 1 ;
else
nB.IsOK = 0 ;
if(minstep[nA.x][nA.y][nB.x][nB.y] > now.step + 1){
que.push(Node(nA , nB , now.step+1)) ;
minstep[nA.x][nA.y][nB.x][nB.y] = now.step + 1 ;
}
}
}
}
return -1 ;
}
int main(){
int T , i , j ;
vector<Point> List ;
cin>>T ;
while(T--){
List.clear() ;
scanf("%d%d" ,&N ,&M) ;
for(i = 1 ; i <= N ; i++)
scanf("%s" , table[i]+1) ;
for(i = 1 ; i <= N ; i++){
for(j = 1 ; j <= M ; j++){
if(table[i][j] == 'B')
List.push_back(Point(i , j , 0)) ;
}
}
int ans =bfs(Node(List[0] , List[1] , 0)) ;
if(ans == -1)
puts("Sorry , sir , my poor program fails to get an answer.") ;
else
printf("%d\n" ,ans) ;
}
return 0 ;
}
HDU2128 Tempter of the Bone II
Problem Description
The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze was changed and the way he came in was lost.He realized that the bone was a trap, and he tried desperately to get out of this maze.
The maze was a rectangle with the sizes of N by M. The maze is made up of a door,many walls and many explosives. Doggie need to reach the door to escape from the tempter. In every second, he could move one block to one of the upper, lower, left or right neighboring blocks. And if the destination is a wall, Doggie need another second explode and a explosive to explode it if he had some explosives. Once he entered a block with explosives,he can take away all of the explosives. Can the poor doggie survive? Please help him.
The maze was a rectangle with the sizes of N by M. The maze is made up of a door,many walls and many explosives. Doggie need to reach the door to escape from the tempter. In every second, he could move one block to one of the upper, lower, left or right neighboring blocks. And if the destination is a wall, Doggie need another second explode and a explosive to explode it if he had some explosives. Once he entered a block with explosives,he can take away all of the explosives. Can the poor doggie survive? Please help him.
Input
The input consists of multiple test cases. The first line of each test case contains two integers N, M,(2 <= N, M <= 8). which denote the sizes of the maze.The next N lines give the maze layout, with each line containing M characters. A character is one of the following:
'X': a block of wall;
'S': the start point of the doggie;
'D': the Door;
'.': an empty block;
'1'--'9':explosives in that block.
Note,initially he had no explosives.
The input is terminated with two 0's. This test case is not to be processed.
'X': a block of wall;
'S': the start point of the doggie;
'D': the Door;
'.': an empty block;
'1'--'9':explosives in that block.
Note,initially he had no explosives.
The input is terminated with two 0's. This test case is not to be processed.
Output
For each test case, print the minimum time the doggie need to escape if the doggie can survive, or -1 otherwise.
Sample Input
4 4 SX.. XX.. .... 1..D 4 4 S.X1 .... ..XX ..XD 0 0
Sample Output
-1 9
typedef long long LL ;
const int Max_N = 9 ;
struct MM{
LL Key ;
LL Exp ;
int ExSize ;
MM(){}
MM(LL a , LL b , int e):Key(a) , Exp(b) , ExSize(e){}
friend bool operator < (const MM &A , const MM &B){
if(A.Key != B.Key)
return A.Key < B.Key ;
if(A.Exp != B.Exp)
return A.Exp < B.Exp ;
return A.ExSize < B.ExSize ;
}
friend bool operator == (const MM &A , const MM &B){
return A.Key == B.Key && A.Exp == B.Exp && A.ExSize == B.ExSize ;
}
};
set< MM > _hash[Max_N][Max_N] ;
struct Node{
int x ;
int y ;
int step ;
int ExSize ;
LL explodes ;
LL key ;
Node(){}
friend bool operator < (const Node &A , const Node &B){
return A.step > B.step ;
}
};
char table[Max_N][Max_N] ;
int minstep[Max_N][Max_N][1000] ;
int N , M ;
int d[4][2] = {{1,0} , {-1 ,0} , {0 , 1} , {0 ,-1}} ;
int cango(int x , int y){
return 1 <= x && x <= N && 1 <= y && y <= M ;
}
int bfs(Node s){
int i , j , k , x , y , idx ;
Node now , next ;
_hash[s.x][s.y].insert(MM(s.key , s.explodes , s.ExSize)) ;
priority_queue<Node> que ;
que.push(s) ;
while(! que.empty()){
now = que.top() ;
que.pop() ;
// printf("%d %d %d %d\n" , now.x , now.y , now.ExSize ,now.step) ;
if(table[now.x][now.y] == 'D')
return now.step ;
for(i = 0 ; i < 4 ; i++){
next.x = now.x + d[i][0] ;
next.y = now.y + d[i][1] ;
if(! cango(next.x , next.y))
continue ;
idx = (next.x - 1) * M + next.y - 1 ;
if(table[next.x][next.y] == 'X'){
next.explodes = now.explodes ;
if(((LL)1<<idx) & now.key){
next.key = now.key ;
next.step = now.step + 1 ;
next.ExSize = now.ExSize ;
}
else{
if(now.ExSize <= 0)
continue ;
next.key = now.key | ((LL)1<<idx) ;
next.ExSize = now.ExSize - 1 ;
next.step = now.step + 2 ;
}
if(_hash[next.x][next.y].find(MM(next.key, next.explodes , next.ExSize)) == _hash[next.x][next.y].end()){
que.push(next) ;
_hash[next.x][next.y].insert(MM(next.key, next.explodes , next.ExSize)) ;
}
}
else if('1' <= table[next.x][next.y] && table[next.x][next.y] <= '9'){
next.key = now.key ;
next.step = now.step + 1 ;
if(((LL)1<<idx) & now.explodes){
next.explodes = now.explodes ;
next.ExSize = now.ExSize ;
}
else{
next.explodes = now.explodes | ((LL)1<<idx) ;
next.ExSize = now.ExSize + table[next.x][next.y] - '0' ;
}
if(_hash[next.x][next.y].find(MM(next.key, next.explodes , next.ExSize)) == _hash[next.x][next.y].end()){
que.push(next) ;
_hash[next.x][next.y].insert(MM(next.key, next.explodes , next.ExSize)) ;
}
}
else{
next.explodes = now.explodes ;
next.step = now.step + 1 ;
next.key = now.key ;
next.ExSize = now.ExSize ;
if(_hash[next.x][next.y].find(MM(next.key, next.explodes , next.ExSize)) == _hash[next.x][next.y].end()){
que.push(next) ;
_hash[next.x][next.y].insert(MM(next.key, next.explodes , next.ExSize)) ;
}
}
}
}
return -1 ;
}
int main(){
int i , j ;
Node start ;
while(cin>>N>>M){
if(N == 0 && M == 0)
break ;
for(i = 1 ; i <= N ; i++)
scanf("%s" ,table[i]+1) ;
for(i = 1 ; i <= N ; i++){
for(j = 1 ; j <= M ; j++){
if(table[i][j] == 'S'){
start.x = i ;
start.y = j ;
}
}
}
for(i = 1 ; i <= N ; i++)
for(j = 1 ; j <= M ; j++)
_hash[i][j].clear() ;
start.step = 0 ;
start.ExSize = 0 ;
start.explodes = (LL)0 ;
start.key = (LL)0 ;
printf("%d\n" , bfs(start)) ;
}
return 0 ;
}