【bfs+优先队列】POJ2049-Finding Nemo

  1 #include<queue>
  2 #include<iostream>
  3 #include<cstdio>
  4 #include<cstring>
  5 using namespace std;
  6 const int MAXN=205;
  7 const int INF=1000000;
  8 struct point
  9 {
 10     int x,y,num;
 11     bool operator < (const point &x) const
 12     {
 13         return num>x.num;
 14         /*优先队列设置为小顶堆*/
 15     }
 16 };
 17 int m,n;
 18 int limitx,limity,littlex,littley,fx,fy;
 19 int map[MAXN][MAXN][4];
 20 int step[MAXN][MAXN];
 21 /*方格的坐标用左下角顶点坐标来记录*/
 22 /*第三维依次代表上下左右*/ 
 23 /*数组内元素的值0：空白 1：墙 2：门*/
 24 int dx[4]={0,0,-1,1};
 25 int dy[4]={1,-1,0,0};
 26 /*d数组要对应上下左右的顺序，方便后续操作*/
 27 int vis[MAXN][MAXN];
 28 float initx,inity;
 29 
 30 void initwall()
 31 {
 32     memset(map,0,sizeof(map));
 33     limitx=limity=-1;
 34     littlex=littley=INF;
 35     int x,y,d,t;
 36     for (int i=0;i<m;i++)
 37     {
 38         cin>>x>>y>>d>>t;
 39         if (x+t>limitx) limitx=x+t;
 40         if (y+t>limity) limity=y+t;
 41         if (x-1<littlex) littlex=x-1;
 42         if (y-1<littley) littley=y-1;
 43         if (d==0)
 45         {
 46           for (int j=x;j<x+t;j++)
 47           {
 49               map[j][y][1]=1;
 50               if (y>=1) map[j][y-1][0]=1;
 51           }
 52         }
 53         else
 54         {
 56             for (int j=y;j<y+t;j++)
 57             {
 58                 map[x][j][2]=1;
 59                 if (x>=1) map[x-1][j][3]=1;
 60             }
 61         }
 62     }
 63 }
 64 
 65 void initdoor()
 66 {
 67     int x,y;
 68     int d;
 69     for (int i=0;i<n;i++)
 70     {
 71         scanf("%d%d%d",&x,&y,&d);
 72         if (x+1>limitx) limitx=x+1;
 73         if (y+1>limity) limity=y+1;
 74         if (x-1<littlex) littlex=x-1;
 75         if (y-1<littley) littley=y-1;
 76         if (d==0)
 77         {
 78                 map[x][y][1]=2;
 79                 if (y>=1) map[x][y-1][0]=2;
 80         }
 81         else
 82         {
 83                 map[x][y][2]=2;
 84                 if (x>=1) map[x-1][y][3]=2;
 85         }
 86     }
 87 }
 88 
 89 int bfs()
 90 {
 91     for (int i=0;i<=limitx;i++)
 92         for (int j=0;j<=limity;j++) step[i][j]=INF;
 93     int ans=INF;
 94     point in,out;
 96     memset(vis,0,sizeof(vis));
 97     step[fx][fy]=0; 
 98     in.x=fx;in.y=fy;in.num=0;
 99     priority_queue<point> que;
100     /*由于队列记录的是经过的门数而非步数，要用优先队列*/
101     que.push(in); 
102     while (!que.empty())
103     {
104         out=que.top();
105         que.pop();
106         if (out.x==0 && out.y==0|| out.x==littlex || out.y==littley || out.x==limitx || out.y==limity)
107         {
108             return out.num;
109         }
110             /*如果到达终点或超出最大范围，直接输出结果*/ 
111         for (int i=0;i<4;i++)
112         {
113             int nowx=out.x+dx[i],nowy=out.y+dy[i],nownum=out.num;
114             if (nowx<0 || nowy<0 || nowx>limitx || nowy>limity) continue;
115             if (map[out.x][out.y][i]==1) continue;
116             /*前进的方向不能被门阻隔*/
117             if (map[out.x][out.y][i]==2) nownum++;
118             /*如果前进的方向是一扇门，则加一*/
119             if (step[nowx][nowy]==INF || nownum<step[nowx][nowy]&& nownum<ans)
120             {
121                 in.x=nowx;in.y=nowy;in.num=nownum;
122                 que.push(in);
123                 step[nowx][nowy]=in.num;
124             }
125         } 
126     }
127     return -1;
128     /*无解时返回-1*/
129 }
130 
131 int main()
132 {
133     while (scanf("%d%d",&m,&n)) 
134     {
135         if (m==n && m==-1) break;
136 
137         initwall();
138         initdoor();
139         cin>>initx>>inity;
140         fx=initx;
141         fy=inity;
142         /*由于方格坐标由左下角定点来表示，强制转换为int型即可*/
143         if (n==0 && m==0||fx>199||fy>199||fx<0||fy<0)
144            cout<<0<<endl;
145         else cout<<bfs()<<endl;
146     }
147     return 0;
148 }