zoukankan      html  css  js  c++  java
  • poj3414Pots(BFS)

    http://poj.org/problem?id=3414

    六种情况 搜一下

    View Code
      1 #include <iostream>
      2 #include<cstdio>
      3 #include<cstring>
      4 using namespace std;
      5 struct node
      6 {
      7     int ca,cb;
      8     int flag,num;
      9     int op,ob;
     10 }q[1000000];
     11 int a,b,c,vis[110][110],num[100000],g;
     12 void put(int x,int y)
     13 {
     14     int yy;
     15     if(y==1)
     16     yy = 2;
     17     else
     18     yy = 1;
     19     if(x==1)
     20         printf("FILL(%d)\n",y);
     21     else
     22     if(x==3)
     23         printf("POUR(%d,%d)\n",y,yy);
     24     else
     25         printf("DROP(%d)\n",y);
     26 }
     27 void bfs()
     28 {
     29     int p = 0,d = 1,na,nb,flag = 0;
     30     memset(vis,0,sizeof(vis));
     31     q[d].ca = 0;
     32     q[d].cb = 0;
     33     q[d].num = 0;
     34     q[d].flag = -1;
     35     g = 0;
     36     while(p!=d)
     37     {
     38         p++;
     39         int ta = q[p].ca;
     40         int tb = q[p].cb;
     41         int tnum = q[p].num;
     42         if(ta==c||tb==c)
     43         {
     44             flag = 1;
     45             break;
     46         }
     47         if(ta!=a)
     48         {
     49             na = a;
     50             nb = tb;
     51             if(!vis[na][nb])
     52             {
     53                 d++;
     54                 q[d].ca = na;
     55                 q[d].cb = nb;
     56                 q[d].flag = p;
     57                 q[d].op = 1;
     58                 q[d].ob = 1;
     59                 q[d].num = tnum+1;
     60                 vis[na][nb] = 1;
     61             }
     62         }
     63         if(tb!=b)
     64         {
     65             nb = b;
     66             na = ta;
     67             if(!vis[na][nb])
     68             {
     69                 d++;
     70                 q[d].ca = na;
     71                 q[d].cb = nb;
     72                 q[d].flag = p;
     73                 q[d].op = 1;
     74                 q[d].ob = 2;
     75                 q[d].num = tnum+1;
     76                 vis[na][nb] = 1;
     77             }
     78         }
     79         if(ta!=0)
     80         {
     81             nb = tb;
     82             na = 0;
     83             if(!vis[na][nb])
     84             {
     85                 d++;
     86                 q[d].ca = na;
     87                 q[d].cb = nb;
     88                 q[d].flag = p;
     89                 q[d].op = 2;
     90                 q[d].ob = 1;
     91                 q[d].num = tnum+1;
     92                 vis[na][nb] = 1;
     93             }
     94         }
     95         if(tb!=0)
     96         {
     97             nb = 0;
     98             na = ta;
     99             if(!vis[na][nb])
    100             {
    101                 d++;
    102                 q[d].ca = na;
    103                 q[d].cb = nb;
    104                 q[d].flag = p;
    105                 q[d].op = 2;
    106                 q[d].ob = 2;
    107                 q[d].num = tnum+1;
    108                 vis[na][nb] = 1;
    109             }
    110         }
    111         if(tb!=b&&ta!=0)
    112         {
    113             if(ta+tb<=b)
    114             {
    115                 na = 0;
    116                 nb = ta+tb;
    117             }
    118             else
    119             {
    120                 nb = b;
    121                 na = ta-(b-tb);
    122             }
    123             if(!vis[na][nb])
    124             {
    125                 d++;
    126                 q[d].ca = na;
    127                 q[d].cb = nb;
    128                 q[d].flag = p;
    129                 q[d].op = 3;
    130                 q[d].ob = 1;
    131                 q[d].num = tnum+1;
    132                 vis[na][nb] = 1;
    133             }
    134         }
    135         if(ta!=a&&tb!=0)
    136         {
    137             if(tb+ta<=a)
    138             {
    139                 nb = 0;
    140                 na = tb+ta;
    141             }
    142             else
    143             {
    144                 na = a;
    145                 nb = tb-(a-ta);
    146             }
    147             if(!vis[na][nb])
    148             {
    149                 d++;
    150                 q[d].ca = na;
    151                 q[d].cb = nb;
    152                 q[d].flag = p;
    153                 q[d].op = 3;
    154                 q[d].ob = 2;
    155                 q[d].num = tnum+1;
    156                 vis[na][nb] = 1;
    157             }
    158         }
    159     }
    160     if(flag)
    161     {
    162         cout<<q[p].num<<endl;
    163         int x = p;
    164         while(x!=1)
    165         {
    166             g++;
    167             num[g] = x;
    168             x = q[x].flag;
    169         }
    170     }
    171     else
    172     puts("impossible");
    173 }
    174 int main()
    175 {
    176     int i;
    177     while(cin>>a>>b>>c)
    178     {
    179         bfs();
    180         for(i = g ; i >= 1; i--)
    181         put(q[num[i]].op,q[num[i]].ob);
    182     }
    183     return 0;
    184 }
  • 相关阅读:
    打造TypeScript的Visual Studio Code开发环境
    Visual Studio Code 使用Git进行版本控制
    每周一书《Oracle 12 c PL(SQL)程序设计终极指南》
    每周一书-《鸟哥的Linux私房菜》获奖公布
    Python黑帽编程 4.1 Sniffer(嗅探器)之数据捕获(上)
    微信公众平台及门户应用
    Bootstrap开发框架
    公用类库使用帮助
    git
    SQL Server 基本函数 与 类型
  • 原文地址:https://www.cnblogs.com/shangyu/p/2877508.html
Copyright © 2011-2022 走看看