3414Pots

不知道是谁出这道题的...坑爹，又是一开始错，没怎么改动，接着又通过了

我的代码ACCEPT

#include "iostream"
#include "queue"
#include "algorithm"
#include "string.h"
using namespace std;
struct Point{
  int a,b,count;
  int list[1000];
};
int min(int a,int b){return a>b?b:a;}
Point tem,ans;
int A,B,C;
char rea[6][20]={{"FILL(2)"},{"FILL(1)"},{"DROP(2)"},{"DROP(1)"},{"POUR(1,2)"},{"POUR(2,1)"}};

void pour(int a){
  int x=tem.a+tem.b;
  if(a==1){
    ans.b=x>B?B:x;
    ans.a=x-ans.b;
  }
  else{
    ans.a=x>A?A:x;
    ans.b=x-ans.a;
  }
}
int dir[6][2]={{0,1},{1,0},{0,-1},{-1,0},{2,3},{3,2}};

int main(){
  int i,j,map[110][110],flag=0;
  cin>>A>>B>>C;
  memset(map,0,sizeof(map));
  Point pt;
  pt.a=0;pt.b=0;pt.count=1;
  queue<Point> q;
  q.push(pt);
  while(!q.empty()){
    tem=q.front();
    q.pop();
    if(map[tem.a][tem.b]==1)continue;
    map[tem.a][tem.b]=1;
    //cout<<tem.a<<' '<<tem.b<<' '<<tem.count<<endl;system("pause");
    for(i=0;i<6;i++){
      if(i==0){ans.a=tem.a;ans.b=B;tem.list[tem.count]=0;}
      else if(i==1){ans.a=A;ans.b=tem.b;tem.list[tem.count]=1;}
      else if(i==2){ans.a=tem.a;ans.b=0;tem.list[tem.count]=2;}
      else if(i==3){ans.a=0;ans.b=tem.b;tem.list[tem.count]=3;}
      else if(i==4){pour(1);tem.list[tem.count]=4;}
      else {pour(-1);tem.list[tem.count]=5;}
      ans.count=tem.count+1;
      for(j=1;j<ans.count;j++){ans.list[j]=tem.list[j];}
      //cout<<ans.a<<' '<<ans.b<<' '<<ans.count<<endl;system("pause");
      q.push(ans);
      if(ans.a==C||ans.b==C){flag=1;goto l1;}
    }
    //cout<<endl<<endl;
  }
l1:if(flag){cout<<ans.count-1<<endl;for(i=1;i<ans.count;i++)cout<<rea[ans.list[i]]<<endl;}
   else cout<<"impossible"<<endl;
}

人家的代码，也不错，不用queue，直接用数组

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#define Min(a,b)a<b?a:b
#define MAX 105
using namespace std;
int a,b,c;
bool vs[MAX][MAX];
char str[6][10]={"FILL(1)","FILL(2)","DROP(1)","DROP(2)",
             "POUR(1,2)","POUR(2,1)"};
struct node
{
    int ope,a,b,step;
    node *pre;
}qu[MAX*MAX];

void print(node now)//输出很有技巧  看一大神的思路
{
    if(now.pre!=NULL)
    {
       print(*now.pre);
       printf("%s
",str[now.ope]);
    }

}

void bfs()//六种操作
{
    int beg=0,end=0;
    qu[beg].a=qu[beg].b=0;
    qu[beg].step=0;
    qu[beg].pre=NULL;
    while(beg<=end)
    {
        node t=qu[beg];
        if(t.a==c||t.b==c)
        {
            printf("%d
",t.step);
            print(t);
            return ;
        }

        if(!vs[a][t.b])
        {
           end++;
           qu[end].ope=0;
           qu[end].a=a;
           qu[end].b=t.b;
           qu[end].step=t.step+1;
           qu[end].pre=&qu[beg];
           vs[a][t.b]=1;
        }
        if(!vs[t.a][b])
        {
            end++;
            qu[end].ope=1;
            qu[end].a=t.a;
            qu[end].b=b;
            qu[end].step=t.step+1;
            qu[end].pre=&qu[beg];
            vs[t.a][b]=1;
        }
        if(!vs[0][t.b])
        {
           end++;
           qu[end].ope=2;
           qu[end].a=0;
           qu[end].b=t.b;
           qu[end].step=t.step+1;
           qu[end].pre=&qu[beg];
           vs[0][t.b]=1;
        }
        if(!vs[t.a][0])
        {
            end++;
            qu[end].ope=3;
            qu[end].a=t.a;
            qu[end].b=0;
            qu[end].step=t.step+1;
            qu[end].pre=&qu[beg];
            vs[t.a][0]=1;
        }
        int min1=Min(t.a,b-t.b);
        if(!vs[t.a-min1][t.b+min1])
        {
            end++;
            qu[end].ope=4;
            qu[end].a=t.a-min1;
            qu[end].b=t.b+min1;
            qu[end].step=t.step+1;
            qu[end].pre=&qu[beg];
            vs[t.a-min1][t.b+min1]=1;
        }
        int min2=Min(a-t.a,t.b);
        if(!vs[t.a+min2][t.b-min2])
        {
           end++;
           qu[end].ope=5;
           qu[end].a=t.a+min2;
           qu[end].b=t.b-min2;
           qu[end].step=t.step+1;
           qu[end].pre=&qu[beg];
           vs[t.a+min2][t.b-min2]=1;
        }
        beg++;
    }
     printf("impossible
");
     return ;
}

int main()
{
    scanf("%d%d%d",&a,&b,&c);
    memset(vs,0,sizeof(vs));
    bfs();

    return 0;
}