hdu4845 BFS

来源：互联网发布：智能家居网络协议编辑：程序博客网时间：2024/06/06 07:00

广度优先搜索，二进制压缩记录当前拥有的钥匙情况，起点和终点确定，跑一遍bfs输出结果即可。

附代码：

#include <iostream>#include <cstring>#include <cstdio>#include <queue>using namespace std;int map[20][20];int f[20][20][20][20];int d[20][20];int mark[20][20][4100];int n,m,P,k;int p[15];int xx[5]={1,-1,0,0};int yy[5]={0,0,1,-1};struct pp{    int x,y;    int a;    int s;    pp()    {        a=0;    }};queue <pp> myqueue;int min(int a,int b){    return a>b?b:a;}void short_path(){    while (!myqueue.empty())    {        pp cur=myqueue.front();        myqueue.pop();        int x=cur.x;        int y=cur.y;        //cout<<x<<" "<<y<<" "<<cur.s<<" "<<cur.a<<endl;        for (int i=0;i<=3;i++)        {            int x1=x+xx[i];            int y1=y+yy[i];            int a=(cur.a|map[x1][y1]);            if (x1>n||y1>m||x1<1||y1<1)                continue ;            int t=f[x][y][x1][y1];            if (t==0)                continue ;            if (t>0)            {                if ((cur.a&p[t])==0)                {                    continue ;                }            }            pp next;            next.x=x1;            next.y=y1;            next.a=a;            next.s=cur.s+1;            d[x1][y1]=min(d[x1][y1],next.s);            if (mark[next.x][next.y][next.a]>next.s||mark[next.x][next.y][next.a]==-1)            {                myqueue.push(next);                mark[next.x][next.y][next.a]=next.s;            }        }    }}int main(){    for (int i=1;i<=10;i++)        p[i]=(1<<i);    while (cin>>n>>m>>P>>k)    {        memset(f,-1,sizeof(f));        memset(map,0,sizeof(map));        memset(mark,-1,sizeof(mark));        for (int i=1;i<20;i++)            for (int t=1;t<20;t++)                d[i][t]=99999999;        for (int i=1;i<=k;i++)        {            int x1,y1,x2,y2,z;            cin>>x1>>y1>>x2>>y2>>z;            f[x1][y1][x2][y2]=f[x2][y2][x1][y1]=z;        }        int s;        cin>>s;        while (s--)        {            int a,b,c;            cin>>a>>b>>c;            map[a][b]|=p[c];        }        pp cur;        cur.x=1;        cur.y=1;        cur.s=0;        d[1][1]=0;        myqueue.push(cur);        short_path();        if (d[n][m]!=99999999)            cout<<d[n][m]<<endl;        else            cout<<-1<<endl;    }}

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