西南民族大学第十二届程序设计竞赛（同步赛） A.逃出机房 (bfs)

• 题意:有来两个人A和B,A追B,A和B每次向上下左右移动一个单位,一共有两扇门,问A是否可以追上B(在门口追上也算合法).
• 题解:当时看题意说在门口也算?就觉得是判断两个人到门口的时间,对他们两个人分别跑bfs,记录他们到每个门口的步数,然后if判断一下即可.
• 代码:

#include <bits/stdc++.h>
#define ll long long
#define fi first
#define se second
#define pb push_back
#define me memset
#define rep(a,b,c) for(int a=b;a<=c;++a)
#define per(a,b,c) for(int a=b;a>=c;--a)
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
using namespace std;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b) {return a/gcd(a,b)*b;}

int n,m;
char s[20][20];
bool st[20][20];
int dx[4]={0,0,-1,1},dy[4]={-1,1,0,0};
int zx,zy,hx,hy;
int ex1,ey1,ex2,ey2;
int res1=INF,res2=INF,res3=INF,res4=INF;

struct misaka{
	int x,y,cnt;
}e;

void bfs1(int x,int y){
	me(st,false,sizeof(st));
	queue<misaka> q;
	int cur=0;
	q.push({x,y,0});

	while(!q.empty()){
		auto tmp=q.front();
		q.pop();

		if(st[tmp.x][tmp.y]) continue;
		st[tmp.x][tmp.y]=true;

		if(s[tmp.x][tmp.y]=='@' && cur<2){
			if(tmp.x==ex1 && tmp.y==ey1){
				res1=tmp.cnt;
			}
			else res2=tmp.cnt;
			cur++;
		}

		rep(i,0,3){
			int tx=tmp.x+dx[i];
			int ty=tmp.y+dy[i];
			if((s[tx][ty]=='*' || s[tx][ty]=='@') && !st[tx][ty]){
				q.push({tx,ty,tmp.cnt+1});
			}
		}
	}
}

void bfs2(int x,int y){
	me(st,false,sizeof(st));
	queue<misaka> q;
	int cur=0;
	q.push({x,y,0});

	while(!q.empty()){
		auto tmp=q.front();
		q.pop();

		if(st[tmp.x][tmp.y]) continue;
		st[tmp.x][tmp.y]=true;

		if(s[tmp.x][tmp.y]=='@' && cur<2){
			if(tmp.x==ex1 && tmp.y==ey1){
				res3=tmp.cnt;
			}
			else res4=tmp.cnt;
			cur++;
		}

		rep(i,0,3){
			int tx=tmp.x+dx[i];
			int ty=tmp.y+dy[i];
			if((s[tx][ty]=='*' || s[tx][ty]=='@') && !st[tx][ty]){
				q.push({tx,ty,tmp.cnt+1});
			}
		}
	}
}

int main() {
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

	cin>>n>>m;

	rep(i,1,n){
		rep(j,1,m){
			cin>>s[i][j];
			if(s[i][j]=='Z') {zx=i,zy=j;}
			else if(s[i][j]=='H') {hx=i,hy=j;}
			else if(s[i][j]=='@'){
				if(!ex1 && !ey1) {ex1=i;ey1=j;}
				else {ex2=i;ey2=j;}
			}
		}
	}

	bfs1(hx,hy);
	bfs2(zx,zy);

	if(res3<res1 || res4<res2) cout<<"give me northeast chicken rice and milk tea TOMORROW!\n";
	else cout<<"give me northeast chicken rice and milk tea!\n";

    return 0;
}