HDU 2102 A计划(三维BFS)

这题太欢乐了......虽然wa了几次，但是想到骑士在两幅图的传送门中传来传去就觉得这骑士太坑了

 

#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
using namespace std;
int n,m,cost,head,tail,ans;
char map[2][15][15];
int sum[2][15][15];
struct node {
    int z,x,y;
} q[11111];
node st, end;

int dirx[4] = {1,-1,0,0};
int diry[4] = {0,0,1,-1};
void init() {
    memset(sum,-1,sizeof(sum));
    ans = 0;
}

bool go(int z,int x,int y) {
    if(x < 0 || x >= n || y < 0 || y >= m) return false;
    if(map[z][x][y] == '*') return false;
    if(sum[z][x][y] != -1) return false;   //
    return true;
}

int bfs() {
    head = 0; tail = 0;
    q[head++] = st;
    sum[st.z][st.x][st.y] = 0;
    while(head != tail) {
        node t = q[tail++];
        node tt;
        if(end.z == t.z && end.x == t.x && end.y == t.y) {
            if(cost >= sum[t.z][t.x][t.y]) {
                return sum[t.z][t.x][t.y];
            }
            else return -1;
        }
        for(int i=0; i<4; i++) {
            tt.z = t.z; tt.x = t.x + dirx[i]; tt.y = t.y + diry[i];
            if(go(tt.z,tt.x,tt.y)) {
                //cout << tt.z << ' ' << tt.x << ' ' << tt.y << endl;
                if(map[tt.z][tt.x][tt.y] == '.' || map[tt.z][tt.x][tt.y] == 'P') {
                    sum[tt.z][tt.x][tt.y] = sum[t.z][t.x][t.y] + 1;
                    q[head++] = tt;
                }
                if(map[tt.z][tt.x][tt.y] == '#' && map[1 - tt.z][tt.x][tt.y] != '*' && map[1 - tt.z][tt.x][tt.y] != '#'){
                    sum[tt.z][tt.x][tt.y] = sum[t.z][t.x][t.y] + 1;
                    sum[1 - tt.z][tt.x][tt.y] = sum[tt.z][tt.x][tt.y];
                    tt.z = 1 - tt.z;
                    q[head++] = tt;
                }
            }
        }
    }
    return -1;
}

int main() {
    int T;
    cin >> T;
    while(T --) {
        init();
        cin >> n >> m >> cost;
        for(int z=0; z<2; z++)
            for(int i=0; i<n; i++)
                for(int j=0; j<m; j++) {
                    cin >> map[z][i][j];
                    if(map[z][i][j] == 'S') {
                        st.z = z;
                        st.x = i;
                        st.y = j;
                    }
                    if(map[z][i][j] == 'P') {
                        end.z = z;
                        end.x = i;
                        end.y = j;
                    }
                }
        if(bfs() == -1) printf("NO\n");
        else printf("YES\n");
    }
    return 0;
}