洛谷1126
凡是涉及到模拟的一般坑也会比较多,在这道题我遇到的坑包括但不限于
- 判断每一步的障碍,并且在判断到障碍后直接跳出,不能继续向前
- 这个球是有体积的,也就是说题目中给我们的只是他边角上的一个点,还需要判断其他4个点是否符合要求
- 注意有可能起始坐标和终点坐标相同
#include<iostream>
#include<utility>
#include<queue>
using namespace std;
typedef long long ll;
#define fi(i,a,b) for(int i = a; i <= b; ++i)
#define fr(i,a,b) for(int i = a; i >= b; --i)
#define x first
#define y second
#define sz(x) ((int)(x).size())
#define pb push_back
using pii = pair<int,int>;
//#define DEBUG
int thing[55][55];
int dir[4] = {0,1,2,3};//表示方向
struct point{
int x,y,step;
int state;
};
int start_x,start_y,end_x,end_y;
char temp;
int res;
int n,m;
bool vis[55][55][4];
queue<point> que;
bool check(int p,int q){
if(thing[p][q]) return false;
if(p + 1 <= n){
if(thing[p+1][q]) return false;
if(q + 1 <= m){
if(thing[p+1][q+1]) return false;
}
else return false;
}
else return false;
if(q + 1 <= m){
if(thing[p][q+1]) return false;
}
else return false;
return true;
}
int bfs(){
queue<point> que;
que.push({start_x,start_y,0,res});
vis[start_x][start_y][res] = true;
while(!que.empty()){
point temp = que.front();
que.pop();
fi(i,1,3){
if(temp.state == 0){
int p = temp.x;
int q = temp.y + i;
if(q < 1 || q > m) continue;
if(!vis[p][q][0]){
if(!check(p,q)){
// vis[p][q][0] = true;
break;
}
// cout << p <<" " << q << " " << temp.step+1;
// cout << endl;
if(p == end_x && q == end_y)
return temp.step + 1;
que.push({p,q,temp.step+1,0});
vis[p][q][0] = true;
}
}
else if(temp.state == 1){
int p = temp.x + i;
int q = temp.y;
if(p < 1 || p > n) continue;
if(!vis[p][q][1]){
if(!check(p,q)){
// vis[p][q][1] = true;
break;
}
// cout << p <<" " << q << " " << temp.step+1;
// cout << endl;
if(p == end_x && q == end_y)
return temp.step + 1;
que.push({p,q,temp.step+1,1});
vis[p][q][1] = true;
}
}
else if(temp.state == 2){
int p = temp.x;
int q = temp.y - i;
if(q < 1 || q > m) continue;
if(!vis[p][q][2]){
if(!check(p,q)){
// vis[p][q][2] = true;
break;
}
// cout << p <<" " << q << " " << temp.step+1;
// cout << endl;
if(p == end_x && q == end_y)
return temp.step + 1;
que.push({p,q,temp.step+1,2});
vis[p][q][2] = true;
}
}
else if(temp.state == 3){
int p = temp.x - i;
int q = temp.y;
if(p < 1 || p > n) continue;
if(!vis[p][q][3]){
if(!check(p,q)){
// vis[p][q][3] = true;
break;
}
// cout << p <<" " << q << " " << temp.step+1;
// cout << endl;
if(p == end_x && q == end_y)
return temp.step + 1;
que.push({p,q,temp.step+1,3});
vis[p][q][3] = true;
}
}
}
fi(i,1,3){
if(i == 2) continue;
int p = (temp.state + i) % 4;
if(!vis[temp.x][temp.y][p])
{
que.push({temp.x,temp.y,temp.step+1,p});
vis[temp.x][temp.y][p] = true;
}
}
}
return -1;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n >> m;
fi(i,1,n) fi(j,1,m)
cin >> thing[i][j];
cin >> start_x >> start_y >> end_x >> end_y;
if(start_x == end_x && start_y == end_y) {
cout << 0 << endl;
return 0;
}
cin >> temp;
if(temp == 'E') res = 0;
else if(temp == 'S') res = 1;
else if(temp == 'W') res = 2;
else res = 3;
cout << bfs() << endl;
#ifdef DEBUG
//freopen(D:\in.txt,r,stdin);
#endif
return 0;
}