# BattleCity

quoc14
c_cpp
18 days ago
4.2 kB
1
Indexable
Never
PrimDijktra
```Level 4
Battle City
Many of us had played the game "Battle city" in our childhood, and some people (like me) even often play it on computer now.

What we are discussing is a simple edition of this game. Given a map that consists of empty spaces, rivers, steel walls and brick walls only. Your task is to get a bonus as soon as possible suppose that no enemies will disturb you (See the following picture).

Your tank can't move through rivers or walls, but it can destroy brick walls by shooting. A brick wall will be turned into empty spaces when you hit it, however, if your shot hit a steel wall, there will be no damage to the wall. In each of your turns, you can choose to move to a neighboring (4 directions, not 8) empty space, or shoot in one of the four directions without a move. The shot will go ahead in that direction, until it go out of the map or hit a wall. If the shot hits a brick wall, the wall will disappear (i.e., in this turn). Well, given the description of a map, the positions of your tank and the target, how many turns will you take at least to arrive there?

Input

﻿ There can be more than one test case in the input. The first line has T, the number of test cases.
Then the totally T test cases are provided in the following lines (T<=10).
The first line of each test case contains two integers M and N (2 <= M, N< = 300). Each of the following M lines contains N uppercase letters, each of which is one of 'Y' (you), 'T' (target), 'S' (steel wall), 'B' (brick wall), 'R' (river) and 'E' (empty space). Both 'Y' and 'T' appear only once.

Output

For each test case, please output the turns you take at least in a separate line. If you can't arrive at the target, output "-1" instead.

﻿

Sample

Input

1
3 4
YBEB
EERE
SSTE

Output
Case #1
8

Case #1
8
Case #2
-1
Case #3
5
Case #4
5
Case #5
13
Case #6
29
Case #7
-1
Case #8
-1
Case #9
228
Case #10
598
#include <iostream>
using namespace std;

int T, n, m;
int xx1, xx2, yy1, yy2;
int d[301][301], selected, mp[301][301];
bool vs[301][301];
int MAX = 2000000000;

int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};

// arr luu cac toa do visit = false nhung d[][] != MAX
int arr[900001][2];
int k;

void dikstra(){
d[xx1][yy1] = 0;
arr[k][0] = xx1, arr[k][1] = yy1;
k++;
while(selected < n*m){
int minV = MAX, minI1 = -1, minI2 = -1, index = 0;

for(int i = 0; i < k; i++){
int x1 = arr[i][0], y1 = arr[i][1];
if(!vs[x1][y1] && d[x1][y1] < minV){
minV = d[x1][y1];
minI1 = x1, minI2 = y1;
index = i;
}
}
if(k>0){
arr[index][0] = arr[k-1][0], arr[index][1] = arr[k-1][1];
k--;
}

if((minI1 == xx2 && minI2 == yy2)){
vs[minI1][minI2] = true;
return;
} else if(minV == MAX) return;
selected++;
vs[minI1][minI2] = true;

for(int j = 0; j < 4; j++){
int x = minI1 + dx[j];
int y = minI2 + dy[j];
if(x >= 0 && y >= 0 && x < n && y < m && !vs[x][y]
&& mp[x][y] > 0 && d[x][y] > d[minI1][minI2] + mp[x][y]){
d[x][y] = d[minI1][minI2] + mp[x][y];
arr[k][0] = x, arr[k][1] = y;
k++;
}
}

}
}

int main(){
freopen("input.txt", "r", stdin);
cin >> T;

for(int tc = 1; tc <= T; tc++){
// Input & Initial
int c = 0;
k = 0, selected = 0;
cin >> n >> m;
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
vs[i][j] = false;
char str;
cin >> str;
if(str == 'Y'){
xx1 = i, yy1 = j;
mp[i][j] = 1;
} else if(str == 'T'){
xx2 = i, yy2 = j;
mp[i][j] = 1;
} else if(str == 'S' || str == 'R'){
mp[i][j] = 0;
vs[i][j] = true;
} else if(str == 'E') mp[i][j] = 1, c++;
else if(str == 'B') mp[i][j] = 2;

d[i][j] = MAX;
}
}

// Solve Problem
/*
if(c == n*m - 2) {
cout << "Case #" << tc << endl << (xx2 - xx1) + (yy2 - yy1) << endl;
continue;
}
*/
dikstra();

// Output
if(vs[xx2][yy2]){
cout << "Case #" << tc << endl << d[xx2][yy2] << endl;
} else{
cout << "Case #" << tc << endl << -1 << endl;
}
}

return 0;
}```