C(clang11) 解法, 执行用时: 3ms, 内存消耗: 384K, 提交时间: 2021-01-29 18:49:16

#include<stdio.h>
int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};
int min=9999;
int x0,y0;
char a[11][11];
int b[11][11]={0};
int x,y;
void duqi(int x1,int y1,int k){
	int f;
	if(x1==x&&y1==y){
	if(k<min)
	min=k;
	return ;
}
for(f=0;f<4;f++){
	x0=x1+dx[f];
	y0=y1+dy[f];
	if(b[y0][x0]==0&&(a[y0][x0]=='*'||a[y0][x0]=='Z'||a[y0][x0]=='H'||a[y0][x0]=='@'))
	{
		
		b[y0][x0]=1;
		duqi(x0,y0,k+1);
		b[y0][x0]=0;
		
	}
}
return ;
}
int main()
{
	int i,j,m,n,x3,y3,x2,y2,h,g;
	int w1,w2;
	scanf("%d%d\n",&m,&n);
	for(i=0;i<m;i++)
	gets(a[i]);
	for(i=0;i<m;i++)
	for(j=0;j<n;j++){
	if(a[i][j]=='Z'){
		x3=j;y3=i;
	}
	if(a[i][j]=='H'){
	x2=j;y2=i;}
	}
	for(i=0;i<m;i++)
	for(j=0;j<n;j++)
	if(a[i][j]=='@'){
	x=j;y=i;
	for(h=0;h<m;h++)
	for(g=0;g<n;g++)
	b[h][g]=0;
	b[y3][x3]=1;
	duqi(x3,y3,0);
	b[y3][x3]=0;
	w1=min;min=9999;
	for(h=0;h<m;h++)
	for(g=0;g<n;g++)
	b[h][g]=0;
	b[y2][x2]=1;
	duqi(x2,y2,0);
	b[y2][x2]=0;
	w2=min;min=9999;
	if(w1<w2)
	{
		printf("give me northeast chicken rice and milk tea TOMORROW!");
		return 0;
	}
	
	}
	printf("give me northeast chicken rice and milk tea!");
}

C++(clang++11) 解法, 执行用时: 6ms, 内存消耗: 376K, 提交时间: 2021-01-01 19:20:35

#include<bits/stdc++.h>
using namespace std;
int n, m, dis[123][123], xz, yz, xh, yh, dir[][2] = {0, 1, 0, -1, -1, 0, 1, 0};
string s[123];
typedef pair<int, int>pii;
queue<pii>q;
int main()
{
	cin >> n >> m;
	for(int i = 0; i < n; i++)
	{
		cin >> s[i];
		for(int j = 0; s[i][j]; j++)
			if(s[i][j] == 'H')
				xh = i, yh = j;
			else if(s[i][j] == 'Z')
				xz = i, yz = j;
	}
	dis[xz][yz] = 2000;
	dis[xh][yh] = 1000;
	q.push(pii(xh, yh));
	q.push(pii(xz, yz));
	while(!q.empty())
	{
		pii cur = q.front();
		q.pop();
		int curx = cur.first, cury = cur.second;
		for(int i = 0; i < 4; i++)
		{
			int newx = curx + dir[i][0];
			int newy = cury + dir[i][1];
			if(newx < 0 || newy < 0 || newx >= n || newy >= m)
				continue;
			if(!dis[newx][newy] && s[newx][newy] != '#')
			{
				dis[newx][newy] = dis[curx][cury] + 1;
				if(s[newx][newy] == '@' && dis[newx][newy] >= 2000)
					return cout << "give me northeast chicken rice and milk tea TOMORROW!", 0;
			}
		}
	}
	cout << "give me northeast chicken rice and milk tea!";
}