• https://vjudge.net/contest/359331#problem/B


    倒水问题,dfs暴力23333

      1 #include <iostream>
      2 #include <cstdio>
      3 #include <queue>
      4 #include <algorithm>
      5 #include <stack>
      6 #include <cstring>
      7 #define inf 2147483647
      8 #define N 1000010
      9 #define p(a) putchar(a)
     10 #define For(i,a,b) for(int i=a;i<=b;++i)
     11 
     12 using namespace std;
     13 int a,b,c;
     14 string k[]={"fill A","fill B","pour A B","pour B A","empty A","empty B"}; 
     15 bool vis[1010][1010],flag;
     16 stack<string>s;
     17 void in(int &x){
     18     int y=1;char c=getchar();x=0;
     19     while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
     20     while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
     21     x*=y;
     22 }
     23 void o(int x){
     24     if(x<0){p('-');x=-x;}
     25     if(x>9)o(x/10);
     26     p(x%10+'0');
     27 }
     28 
     29 void dfs(int x,int y){
     30     if(x==c||y==c){
     31         flag=1;
     32         s.push("success");
     33         return;
     34     }
     35     if(!vis[a][y]){
     36         vis[a][y]=1;
     37         dfs(a,y);
     38         if(flag){
     39             s.push(k[0]);
     40             return;
     41         }
     42         vis[a][y]=0;
     43     }
     44     if(!vis[x][b]){
     45         vis[x][b]=1;
     46         dfs(x,b);
     47         if(flag){
     48             s.push(k[1]);
     49             return;
     50         }
     51         vis[x][b]=0;
     52     }
     53     if(x+y<=b){
     54         if(!vis[0][x+y]){
     55             vis[0][x+y]=1;
     56             dfs(0,x+y);
     57             if(flag){
     58                 s.push(k[2]);
     59                 return;
     60             }
     61             vis[0][x+y]=0;
     62         }
     63     }
     64     else{
     65         if(!vis[x+y-b][b]){
     66             vis[x+y-b][b]=1;
     67             dfs(x+y-b,b);
     68             if(flag){
     69                 s.push(k[2]);
     70                 return;
     71             }
     72             vis[x+y-b][b]=0;
     73         }
     74     }
     75     if(x+y<=a){
     76         if(!vis[x+y][0]){
     77             vis[x+y][0]=1;
     78             dfs(x+y,0);
     79             if(flag){
     80                 s.push(k[3]);
     81                 return;
     82             }
     83             vis[x+y][0]=0;
     84         }
     85     }
     86     else{
     87         if(!vis[a][x+y-a]){
     88             vis[a][x+y-a]=1;
     89             dfs(a,x+y-a);
     90             if(flag){
     91                 s.push(k[3]);
     92                 return;
     93             }
     94             vis[a][x+y-a]=0;
     95         }
     96     }
     97     if(!vis[0][y]){
     98         vis[0][y]=1;
     99         dfs(0,y);
    100         if(flag){
    101             s.push(k[4]);
    102             return;
    103         }
    104         vis[0][y]=0;
    105     }
    106     if(!vis[x][0]){
    107         vis[x][0]=1;
    108         dfs(x,0);
    109         if(flag){
    110             s.push(k[5]);
    111             return;
    112         }
    113         vis[x][0]=0;
    114     }
    115 }
    116 
    117 signed main(){
    118     while(cin>>a>>b>>c){
    119         memset(vis,0,sizeof(vis));
    120         flag=0;
    121         dfs(0,0);
    122         while(!s.empty()){
    123             cout<<s.top()<<'
    ';s.pop();
    124         }
    125     }
    126     return 0;
    127 }
  • 相关阅读:
    调用API接口,查询手机号码归属地(2)
    调用API接口,查询手机号码归属地(1)
    F
    Icebound and Sequence(非互质逆元 快速乘法)or(矩阵快速幂)
    ProblemC、小花梨判连通(dfs)+想法stl
    cwb个人练习
    Fire Net (二分图匹配 匈牙利算法模板)
    Fire Net HDU
    Going Home POJ
    Two Sequences (二分+二进制) (好题)
  • 原文地址:https://www.cnblogs.com/war1111/p/12372224.html
Copyright © 2020-2023  润新知