1 /* 2 bfs搜索!要注意的是点与点的权值是不一样的哦! 3 空地到空地的步数是1, 空地到墙的步数是2(轰一炮+移过去) 4 所以用到优先队列进行对当前节点步数的更新! 5 */ 6 #include<iostream> 7 #include<queue> 8 #include<cstring> 9 #include<algorithm> 10 #include<cstdio> 11 using namespace std; 12 13 int n, m; 14 char map[305][305]; 15 16 struct node{ 17 int x, y; 18 int step; 19 node(){} 20 node(int x, int y, int step){ 21 this->x=x; 22 this->y=y; 23 this->step=step; 24 } 25 }; 26 int dir[4][2]={0, 1, 1, 0, -1, 0, 0, -1}; 27 28 bool operator >(node a, node b){ 29 return a.step > b.step; 30 } 31 32 priority_queue<node, vector<node>, greater<node> >q; 33 34 bool bfs(){ 35 while(!q.empty()){ 36 node cur=q.top(); 37 q.pop(); 38 if(map[cur.x][cur.y]==‘T‘){ 39 cout<<cur.step<<endl; 40 return true; 41 } 42 int xx, yy; 43 for(int i=0; i<4; ++i){ 44 xx=cur.x+dir[i][0]; 45 yy=cur.y+dir[i][1]; 46 if(map[xx][yy]==‘R‘ || map[xx][yy]==‘S‘) continue; 47 else if(map[xx][yy]==‘T‘){ 48 cout<<cur.step+1<<endl; 49 return true; 50 } 51 else if(map[xx][yy]==‘B‘) 52 q.push(node(xx, yy, cur.step+2)); 53 else 54 q.push(node(xx, yy, cur.step+1)); 55 56 map[xx][yy]=‘R‘; 57 } 58 } 59 return false; 60 } 61 62 int main(){ 63 while(cin>>n>>m && (n || m)){ 64 for(int i=1; i<=n; ++i){ 65 cin>>(map[i]+1); 66 map[i][0]=map[i][m+1]=‘R‘; 67 for(int j=1; j<=m; ++j){ 68 if(map[i][j]==‘Y‘){ 69 q.push(node(i, j, 0)); 70 map[i][j]=‘R‘; 71 } 72 map[0][j]=map[n+1][j]=‘R‘; 73 } 74 } 75 if(!bfs()) 76 cout<<"-1"<<endl; 77 while(!q.empty()) q.pop(); 78 } 79 return 0; 80 }
1 /* 2 将map[i][j]映射到 i*m+j的节点上,建立节点与节点之间的权值的关系! 3 B->B的权值为1, E->B的权值为2, S<->... R<->... 的权值为INF(也就是没有边存在) 4 在注意一点就是B->E的权值是 1,因为如果到B了,说明炮弹已经将墙轰掉了! 5 6 建立好图之后,那么就是求源点到终点的最短的距离了! 7 这里采用的spfa算法! 8 */ 9 10 #include<iostream> 11 #include<cstdio> 12 #include<cstring> 13 #include<algorithm> 14 #include<vector> 15 #include<queue> 16 #define N 90010 17 #define INF 0x3f3f3f3f 18 using namespace std; 19 struct node{ 20 int to; 21 int dist; 22 node(){} 23 24 node(int to, int dist){ 25 this->to=to; 26 this->dist=dist; 27 } 28 }; 29 vector<node>g[N]; 30 int vis[N], d[N]; 31 char map[305][305]; 32 int dir[4][2]={0, 1, 1, 0, -1, 0, 0, -1}; 33 int ss, tt; 34 int n, m; 35 queue<int>q; 36 bool spfa(){ 37 q.push(ss); 38 memset(vis, 0, sizeof(vis)); 39 vis[ss]=1; 40 memset(d, 0x3f, sizeof(d)); 41 d[ss]=0; 42 while(!q.empty()){ 43 int u=q.front(); q.pop(); 44 vis[u]=0; 45 int len=g[u].size(); 46 for(int i=0; i<len; ++i){ 47 int v=g[u][i].to; 48 if(d[v] > d[u] + g[u][i].dist){ 49 d[v] = d[u] + g[u][i].dist; 50 51 if(!vis[v]){ 52 q.push(v); 53 vis[v]=1; 54 } 55 } 56 } 57 } 58 if(d[tt]==INF) return false; 59 return true; 60 } 61 62 int main(){ 63 while(cin>>n>>m && (n||m)){ 64 for(int i=0; i<n; ++i) 65 cin>>map[i]; 66 for(int i=0; i<n; ++i) 67 for(int j=0; j<m; ++j){ 68 int from=i*m+j; 69 if(map[i][j]==‘Y‘) ss=from; 70 else if(map[i][j]==‘T‘) tt=from; 71 else if(map[i][j]==‘R‘ || map[i][j]==‘S‘) continue; 72 for(int k=0; k<4; ++k){ 73 int x=i+dir[k][1]; 74 int y=j+dir[k][0]; 75 if(x<0 || x>=n || y<0 || y>=m) continue; 76 if(map[x][y]==‘R‘ || map[x][y]==‘S‘) continue; 77 78 int to = x*m+y, dist=1; 79 if(map[i][j]==‘B‘ || map[x][y]==‘B‘) dist=2; 80 if(map[i][j]==‘B‘ && map[x][y]!=‘B‘) dist=1; 81 g[from].push_back(node(to, dist)); 82 83 } 84 } 85 if(!spfa()) 86 cout<<"-1"<<endl; 87 else cout<<d[tt]<<endl; 88 for(int i=0; i<n*m; ++i) 89 g[i].clear(); 90 } 91 return 0; 92 }
POJ 2312Battle City(BFS-priority_queue 或者是建图spfa),布布扣,bubuko.com
时间: 2024-10-10 15:52:03