思路：求边权为0,1的最短路
代码：
#include <iostream>#include <cstring>#include <algorithm>#include <deque>using namespace std;const int N = 1010;#define x first#define y secondtypedef pair<int, int> PII;bool g[N][N],st[N][N];int dis[N][N];int bfs(int sx,int sy){deque<PII> q;q.push_back({sx,sy});int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};memset(dis,0x3f,sizeof dis);dis[sx][sy]=0;while(q.size()){auto t=q.front();q.pop_front();if(st[t.x][t.y]) continue;st[t.x][t.y]=true;if(!t.x&&!t.y) break;for(int i=0;i<4;i++){int x=t.x+dx[i];int y=t.y+dy[i];if(x>=0&&x<N&&y>=0&&y<N){int w=0;if(g[x][y]) w=1;if(dis[x][y]>dis[t.x][t.y]+w){dis[x][y]=dis[t.x][t.y]+w;if(!w) q.push_front({x,y});else q.push_back({x,y});}}}}return dis[0][0];}int main(){int n,sx,sy;scanf("%d%d%d", &n, &sx ,&sy);while (n -- ){int x,y;scanf("%d%d", &x, &y);g[x][y]=true;}printf("%d\n",bfs(sx,sy));}
