UVA 11624 UVA 10047 两道用 BFS进行最短路搜索的题

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
struct node{
    int ft;
    int sta;
}flo[1010][1010];
int vis[1010][1010];
struct person{
    int x,y,t,fx,fy;
};
int R,C;
int dir[][2]={{0,1},{0,-1},{1,0},{-1,0}};
typedef pair<int,int> pill;
queue<pill> v;
queue<person> q;
void init()
{
    memset(vis,0,sizeof vis);
    while (!v.empty()){
        pill x=v.front();
        v.pop();
        for (int i=0;i<4;i++){
            int nx=x.first+dir[i][0];
            int ny=x.second+dir[i][1];
            int tmp=flo[x.first][x.second].ft+1;;
            if (nx<0 || ny<0 || nx>=R || ny>=C) continue;
            if (flo[nx][ny].ft>=0 && flo[nx][ny].ft<=tmp || flo[nx][ny].sta==0) continue;
            flo[nx][ny].ft=flo[x.first][x.second].ft+1;
            pill b=make_pair(nx,ny);
            if (!vis[nx][ny])
             v.push(b);
            vis[nx][ny]=1;
        }
    }
}
int bfs(person x)
{
    memset(vis,0,sizeof vis);
    while (!q.empty()) q.pop();
    q.push(x);
    int s=1<<30;
    while (!q.empty()){
        person u=q.front();
        q.pop();
        if (u.t>=s) continue;
        if (u.x==0 || u.y==0 || u.x==R-1 || u.y==C-1) {s=u.t;break;}
        for (int i=0;i<4;i++){
         int xx=u.x+dir[i][0];
         int yy=u.y+dir[i][1];
         if (xx<0 || yy<0 || xx>=R || yy>=C)  continue;
         if (xx==u.fx && yy==u.fy) continue;
         if (flo[xx][yy].sta!=1 || flo[xx][yy].ft>=0 && flo[xx][yy].ft<=u.t+1) continue;
         person b=(person){xx,yy,u.t+1,u.x,u.y};
         if (!vis[xx][yy]) q.push(b);
         vis[xx][yy]=1;
        }
    }
    return s;
}
int main()
{
    int t,sx,sy;char ch;
    scanf("%d",&t);
    while (t--){
        while (!v.empty()) v.pop();
        scanf("%d%d",&R,&C);
        getchar();
        for (int i=0;i<R;i++){
          for (int j=0;j<C;j++){
                scanf("%c",&ch);
                //cout<<ch<<endl;
                if (ch==‘.‘) {flo[i][j].sta=1;flo[i][j].ft=-1;}
                else if (ch==‘#‘){flo[i][j].sta=0;flo[i][j].ft=-1;}
                else if (ch==‘F‘){
                    flo[i][j].sta=flo[i][j].ft=0;
                    pill a;a.first=i;a.second=j;v.push(a);
                }
                else if (ch==‘J‘) sx=i,sy=j;
          }
          getchar();
        }
        init();
        person a=(person){sx,sy,0,-1,-1};
        int ans=bfs(a);
        if (ans<(1<<30)) printf("%d\n",ans+1);
        else  puts("IMPOSSIBLE");
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int R,C;
int vis[30][30][5][5];
int mat[30][30];
int sx,sy,ex,ey;
int dir[][2]={{-1,0},{0,1},{1,0},{0,-1}};
struct t1{
    int x,y,d,c,t;
};
int bfs(t1 a)
{
    queue<t1> q;
    q.push(a);
    memset(vis,0,sizeof vis);
    while (!q.empty()){
        t1 u=q.front();
        q.pop();
        if (u.x==ex && u.y==ey && u.c==0){
            //cout<<" pass "<<u.x<<" "<<u.y<<endl;
            return u.t;
        }
        vis[u.x][u.y][u.d][u.c]=1;
        int nd=u.d+1;
        if (nd>3) nd=0;
        t1 nx=u;
        nx.d=nd;
        nx.t=u.t+1;
        if (!vis[nx.x][nx.y][nx.d][nx.c]) q.push(nx);
        vis[nx.x][nx.y][nx.d][nx.c]=1;
        nd=u.d-1;
        if (nd<0) nd=3;
        nx=u; nx.d=nd; nx.t=u.t+1;
        if (!vis[nx.x][nx.y][nx.d][nx.c]) q.push(nx);
        vis[nx.x][nx.y][nx.d][nx.c]=1;
        int xx=u.x+dir[u.d][0];
        int yy=u.y+dir[u.d][1];
        if (xx<0 || yy<0 || xx>=R || yy>=C) continue;
        if (mat[xx][yy]==0) continue;
        int nc=u.c+1;
        if (nc>4) nc=0;
        t1 b=(t1){xx,yy,u.d,nc,u.t+1};
        if (!vis[b.x][b.y][b.d][b.c]) q.push(b);
        vis[b.x][b.y][b.d][b.c]=1;
    }
    return -1;
}
int main()
{
    char ch;
    int kase=0;
    while (scanf("%d%d",&R,&C)){
        if (R==0) break;
        getchar();
        memset(mat,0,sizeof mat);
        for (int i=0;i<R;i++){
            for (int j=0;j<C;j++){
                ch=getchar();
                if (ch!=‘#‘) mat[i][j]=1;
                if (ch==‘S‘) sx=i,sy=j;
                if (ch==‘T‘) ex=i,ey=j;
            }
            getchar();
        }
        //cout<<ex<<" exy "<<ey<<endl;
        t1 a=(t1){sx,sy,0,0,0};
       int ans=bfs(a);
       if (kase) puts("");
       printf("Case #%d\n",++kase);
       if (ans==-1)puts("destination not reachable");
       else printf("minimum time = %d sec\n",ans);
    }
    return 0;
}