题解

#include<bits/stdc++.h>
using namespace std;

int n,m,k;
bool vis[205][205]; //标记是否出现过
struct node
{
    int a,b; //当前第一个桶和第二个桶的储水量
    vector<string> opt; //操作序列
};
vector<string> qstr(vector<string> opt,string type,int a,int b) //转换成操作序列，注意fill和drop操作不需要第四个参数
{
    string s="",t="";
    s+=a+'0'; t+=b+'0';
    if (type=="FILL" || type=="DROP")
        opt.push_back(type+"("+s+")");
    else
        opt.push_back("POUR("+s+","+t+")");
    return opt;
}
void bfs()
{
    queue<node> q;
    node tmp; tmp.a=0,tmp.b=0;
    q.push(tmp);
    vis[0][0]=1;
    while (!q.empty())
    {
        int na=q.front().a,nb=q.front().b;
        vector<string> nopt=q.front().opt;
        if (na==k || nb==k) //找到了
        {
            cout << nopt.size() << endl;
            for (auto v:nopt)
                cout << v << endl;
            return;
        }
        q.pop();
        if (vis[n][nb]==0 && na<n) //fill(1,2)，检查na小于n，避免无效操作
        {
            vis[n][nb]=1;
            q.push(node{n,nb,qstr(nopt,"FILL",1,2)});
        }
        if (vis[na][m]==0 && nb<m)
        {
            vis[na][m]=1;
            q.push(node{na,m,qstr(nopt,"FILL",2,1)});
        }
        if (vis[0][nb]==0 && na>0)
        {
            vis[0][nb]=1;
            q.push(node{0,nb,qstr(nopt,"DROP",1,2)});
        }
        if (vis[na][0]==0 && nb>0)
        {
            vis[na][0]=1;
            q.push(node{na,0,qstr(nopt,"DROP",2,1)});
        }
        if (na>0 && nb<m) //倒水
        {
            int aa=na,bb=nb;
            bb+=aa;
            aa=0;
            if (bb>m) aa+=bb-m,bb=m;
            if (vis[aa][bb]==0)
            {
                vis[aa][bb]=1;
                q.push(node{aa,bb,qstr(nopt,"POUR",1,2)});
            }
        }
        if (nb>0 && na<n)
        {
            int aa=na,bb=nb;
            aa+=bb;
            bb=0;
            if (aa>n) bb+=aa-n,aa=n;
            if (vis[aa][bb]==0)
            {
                vis[aa][bb]=1;
                q.push(node{aa,bb,qstr(nopt,"POUR",2,1)});
            }
        }
    }
    cout << "impossible"; //所有状态遍历完了还没找到就impossible
    return;
}
int main()
{
    cin >> n >> m >> k;
    bfs();
	return 0;
}