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?
- 輸入
- The input consists of several test cases. 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. A test case of M = N = 0 indicates the end of input, and should not be processed.
- 輸出
- 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.
- 樣例輸入
-
3 4 YBEB EERE SSTE 0 0
- 樣例輸出
-
8
-
這道題需要用到優先隊列，每一次需要對隊列排序，把短的放前邊，難點就在此。接着用寬搜就能做出來。E代表走一步，B代表走兩步。
-
#include <iostream>
#include <queue>
using namespace std;
struct node {
int x, y;
int step;
};
priority_queue<node> squeue;
char map[320][320];
int bnext[4][2] = { 1,0,-1,0,0,-1,0,1 };
bool operator< (const node &a,const node &b)
{
return a.step > b.step;
}
int main()
{
int n, m;
while (true)
{
int i, j;
node start, s, sk;
int sum = 0;
cin >> n >> m;
if (!n&&!m)
break;
while (!squeue.empty())
{
squeue.pop();
}
for (i = 1; i <= n; i++)
for (j = 1; j <= m; j++)
{
cin >> map[i][j];
if (map[i][j] == 'Y')
{
start.x = i;
start.y = j;
start.step = 0;
}
}
bool flag = false;
squeue.push(start);
while (!squeue.empty())
{
s = squeue.top();
squeue.pop();
for (i = 0; i <= 3; i++)
{
sk.x = s.x + bnext[i][0];
sk.y = s.y + bnext[i][1];
//sk.step = s.step;
if (sk.x >= 1 && sk.x <= n&&sk.y >= 1 && sk.y <= m)
{
//cout << sk.x << " " << sk.y << " " <<sk.step<<" "<< endl;
if (map[sk.x][sk.y] == 'T')
{
sum = s.step + 1;
flag = true;
break;
}
else if (map[sk.x][sk.y] == 'B')
{
sk.step = s.step + 2;
squeue.push(sk);
map[sk.x][sk.y] = 'S';
}
else if (map[sk.x][sk.y] == 'E')
{
sk.step = s.step + 1;
squeue.push(sk);
map[sk.x][sk.y] = 'S';
}
}
}
if (flag)
break;
}
if (flag)
cout << sum << endl;
else
cout << "-1" << endl;
}
}