bzoj 1687: [Usaco2005 Open]Navigating the City 城市交通 bfs

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bfs时，对于每个点，显然最多只能访问一次，并且只要访问到了就一定是最短路径

记录一下到达每个点的步数，最后从终点每次找周围步数比它小1的点，就可以找到我们需要的那个路径

已经知道路径了，反着搞回去就可以了，每次判断一下行走方向是否改变就好

懒得反着记录答案，所以用了个双向队列，真是方便啊

代码：

#include<iostream>#include<cstdio>#include<queue>#include<deque>#include<vector>#include<algorithm>#include<cstring>#define inf 999999999#define ll long longusing namespace std;struct point{int x,y;friend bool operator == (point a,point b){return (a.x==b.x && a.y==b.y);}};struct node{char way;int num;};point s,e;int n,m;int a[110][110];int dis[110][110];queue<point>q;int tox[4]={-1,0,1,0};int toy[4]={0,1,0,-1};deque<node>ans;int main(){scanf("%d%d\n",&n,&m);n=n*2-1;m=m*2-1;for(int i=1; i<=n; i++){for(int j=1; j<=m; j++){dis[i][j]=-1;char c;c=getchar();if(c=='.')continue;else if(c=='+')a[i][j]=2;else if(c=='S')s.x=i,s.y=j,a[i][j]=2;else if(c=='E')e.x=i,e.y=j,a[i][j]=2;else a[i][j]=1;}getchar();}dis[s.x][s.y]=0;point t;t=s;q.push(t);while(!q.empty()){t=q.front();q.pop();if(t==e)break;for(int i=0; i<4; i++){point t1=t;t1.x+=tox[i];t1.y+=toy[i];if(t1.x<1 || t1.x>n || t1.y<1 || t1.y>m || a[t1.x][t1.y]==0 || dis[t1.x][t1.y]!=-1)continue;dis[t1.x][t1.y]=dis[t.x][t.y]+1;q.push(t1);}}int last=-1;int sum=0;t=e;while(1){if(t==s){node tt;if(last==0)tt.way='S';if(last==1)tt.way='W';if(last==2)tt.way='N';if(last==3)tt.way='E';tt.num=sum;ans.push_front(tt);break;}int to;if(a[t.x][t.y]==2)sum++;for(int i=0; i<4; i++){point t1=t;t1.x+=tox[i];t1.y+=toy[i];if(t1.x<1 || t1.x>n || t1.y<1 || t1.y>m || a[t1.x][t1.y]==0 || dis[t1.x][t1.y]!=dis[t.x][t.y]-1)continue;to=i;break;}if(last!=to){if(last!=-1){node tt;if(last==0)tt.way='S';if(last==1)tt.way='W';if(last==2)tt.way='N';if(last==3)tt.way='E';tt.num=sum-1;ans.push_front(tt);sum=1;}last=to;}t.x+=tox[to];t.y+=toy[to];}for(int i=0; i<ans.size(); i++){printf("%c %d\n",ans[i].way,ans[i].num);}return 0;}