UVA 11624 UVA 10047 两道用 BFS进行最短路搜索的题

很少用bfs进行最短路搜索，实际BFS有时候挺方便得，省去了建图以及复杂度也降低了O(N*M)；

UVA 11624 写的比较挫

#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;

}

UVA 10047

#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;

}