poj2312Battle City BFS

题意: M行N列矩阵, 'Y'表示开始位置, 'T'表示目标位置, 从开始位置到目标位置至少需要走多少步,其中, 'S', 'R'表示不能走, 'B' 花费为2, 'E'花费为1.

思路:纯 BFS.

很久没写BFS, 刚写居然不知道从何下手...

#include <iostream>
#include <queue>
using namespace std;

struct postion
{
    int i, j;
    postion operator+(postion p)
    {
        postion t;
        t.i = this->i+ p.i;
        t.j = this->j+ p.j;
        return t;
    }
};

postion dir[] = {1,0, -1,0, 0,1, 0,-1};

int step[305][305];
//bool used[305][305];
queue<postion>que;
char map[305][305];
postion start, target;
int M, N;

int main()
{
    int  i, j;
    while (cin>>M>>N && M+N)
    {
        for (i=0; i<M; i++)
            for (j=0; j<N; j++)
            {
                cin>>map[i][j];
                step[i][j] = INT_MAX;
                if (map[i][j] == 'Y')
                    start.i = i, start.j = j;
                else if (map[i][j] == 'T')
                    target.i = i, target.j = j;
            }
            //memset(used, false, sizeof(used));
            //memset(step, 0, sizeof(step));
            //used[start.i][start.j] = true;
            step[start.i][start.j] = 0;
            while (!que.empty())que.pop();
            que.push(start);
            postion tmp, t;
            int sp;
            while (!que.empty())
            {
                tmp = que.front();
                que.pop();
                for (i=0; i<4; i++)
                {
                    t = tmp + dir[i];
                    if (t.i<0 || t.i>=M || t.j<0 || t.j>=N || map[t.i][t.j] == 'R' || map[t.i][t.j] == 'S')
                        continue;
                    if (map[t.i][t.j] == 'E')
                        sp = 1;
                    if (map[t.i][t.j] == 'B')
                        sp = 2;
                    if (step[t.i][t.j] > step[tmp.i][tmp.j] + sp)
                    {
                        step[t.i][t.j] = step[tmp.i][tmp.j] + sp;
                        que.push(t);
                    }
                }
            }
            if (step[target.i][target.j] == INT_MAX)
                step[target.i][target.j] = -1;
            cout<<step[target.i][target.j]<<endl;
    }
    return 0;
}