题意：有两个固定大小的容器，通过对容器内水的操作，使得至少一个容器内的液体量达到规定的量，求出操作数最少的操作步骤

我写的比较繁琐，将各种操作封装在了结构体里，然后BFS

代码：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <cstdio>
#include <vector>
#include <iomanip>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#define ll long long
#define mod 10000000007
#define mem(a) memset(a,0,sizeof(a))using namespace std;typedef pair <int,int> pii;
const int maxn = 10000+5 , inf = 0x3f3f3f3f;
int C;
int a[3];
bool vis[maxn][maxn];
struct Node{int i,j,id,f,flag;int A,B;Node(int ii=0,int jj=0,int ff=0,int idd=0,int flagg=0,int AA=0,int BB=0):i(ii),j(jj),f(ff),id(idd),flag(flagg),A(AA),B(BB){};void FILL(){if(i==1) A=a[1];else B=a[2];}void DROP(){if(i==1) A=0;else B=0;}void POUR(){if(i==1&&j==2){if(A>a[2]-B){A-=(a[2]-B);B=a[2];}else{B+=A;A=0;}}if(i==2&&j==1){if(B>a[1]-A){B-=(a[1]-A);A=a[1];}else{A+=B;B=0;}}}void print(){if(j==0){if(flag==1){if(i==1) cout<<"FILL(1)"<<endl;if(i==2) cout<<"FILL(2)"<<endl;}if(flag==3){if(i==1) cout<<"DROP(1)"<<endl;if(i==2) cout<<"DROP(2)"<<endl;}}else{if(i==1&&j==2) cout<<"POUR(1,2)"<<endl;if(i==2&&j==1) cout<<"POUR(2,1)"<<endl;}}
}oper[maxn];bool judge(Node now){if(now.A==C||now.B==C) return true;return false;
}
vector<Node>path;
int find_path(Node u){while(u.f!=u.id){path.push_back(u);u = oper[u.f];}path.push_back(u);cout<<path.size()<<endl;for(int i=path.size()-1;i>=0;i--)path[i].print();
}void bfs(){vis[0][0] = true;Node tmp1(1,0,0,0,1,0,0);oper[0]=tmp1;Node tmp2(2,0,1,1,1,0,0);oper[1]=tmp2;Node tmp3(1,2,2,2,2,0,0);oper[2]=tmp3;Node tmp4(2,1,3,3,2,0,0);oper[3]=tmp4;Node tmp5(1,0,4,4,3,0,0);oper[4]=tmp5;Node tmp6(2,0,5,5,3,0,0);oper[5]=tmp6;queue<Node>q;q.push(tmp1);q.push(tmp2);q.push(tmp3);q.push(tmp4);q.push(tmp5);q.push(tmp6);int cnt = 6;while(!q.empty()){Node now = q.front();q.pop();if(now.j==0){if(now.flag==1) now.FILL();if(now.flag==3) now.DROP();}else now.POUR();if(judge(now)){find_path(now);return ;}if(!vis[now.A][now.B]){vis[now.A][now.B] = true;Node tmp1(1,0,now.id,cnt,1,now.A,now.B);oper[cnt++]=tmp1;q.push(tmp1);Node tmp2(2,0,now.id,cnt,1,now.A,now.B);oper[cnt++]=tmp2;q.push(tmp2);Node tmp3(1,2,now.id,cnt,2,now.A,now.B);oper[cnt++]=tmp3;q.push(tmp3);Node tmp4(2,1,now.id,cnt,2,now.A,now.B);oper[cnt++]=tmp4;q.push(tmp4);Node tmp5(1,0,now.id,cnt,3,now.A,now.B);oper[cnt++]=tmp5;q.push(tmp5);Node tmp6(2,0,now.id,cnt,3,now.A,now.B);oper[cnt++]=tmp6;q.push(tmp6);}}cout<<"impossible"<<endl;
}int main(){//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);scanf("%d%d%d",&a[1],&a[2],&C);bfs();
}