POJ1606 Jugs
作者:互联网
解法一:\(BFS\)
每次扩展的状态有\(6\)个。宽搜多写写还是有好处的。
const int N=110;
struct Node
{
int ca,cb;
int dist;
string path;
};
bool vis[N][N];
string op[]={"FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)"};
int ca,cb,target;
bool bfs()
{
queue<Node> q;
q.push({0,0,0,""});
while(q.size())
{
Node t=q.front();
q.pop();
if(t.ca == target || t.cb == target)
{
cout<<t.dist<<endl;
for(int i=0;i<t.path.size();i++)
{
int k=t.path[i]-'0';
cout<<op[k]<<endl;
}
return true;
}
if(t.ca < ca && !vis[ca][t.cb])
{
vis[ca][t.cb]=true;
q.push({ca,t.cb,t.dist+1,t.path+'0'});
}
if(t.cb < cb && !vis[t.ca][cb])
{
vis[t.ca][cb]=true;
q.push({t.ca,cb,t.dist+1,t.path+'1'});
}
if(t.ca && !vis[0][t.cb])
{
vis[0][t.cb]=true;
q.push({0,t.cb,t.dist+1,t.path+'2'});
}
if(t.cb && !vis[t.ca][0])
{
vis[t.ca][0]=true;
q.push({t.ca,0,t.dist+1,t.path+'3'});
}
if(t.ca && t.cb < cb)
{
int na,nb;
if(t.ca + t.cb > cb) na=t.ca+t.cb-cb, nb=cb;
else na=0, nb=t.ca+t.cb;
if(!vis[na][nb])
{
vis[na][nb]=true;
q.push({na,nb,t.dist+1,t.path+'4'});
}
}
if(t.cb && t.ca < ca)
{
int na,nb;
if(t.ca + t.cb > ca) na=ca, nb=t.ca+t.cb-ca;
else na=t.ca+t.cb, nb=0;
if(!vis[na][nb])
{
vis[na][nb]=true;
q.push({na,nb,t.dist+1,t.path+'5'});
}
}
}
return false;
}
int main()
{
cin>>ca>>cb>>target;
if(!bfs()) puts("impossible");
//system("pause");
}
标签:Jugs,int,cb,nb,vis,na,POJ1606,ca 来源: https://www.cnblogs.com/fxh0707/p/14443465.html