BFS POJ 3414 Pots

 

题目传送门

  1 /*
  2     BFS：六种情况讨论一下，BFS轻松解决
  3     起初我看有人用DFS，我写了一遍，TLE。。还是用BFS，结果特判时出错，逗了好长时间
  4     看别人的代码简直是受罪，还好自己终于发现自己代码的小错误：)
  5 */
  6 /************************************************
  7 Author        :Running_Time
  8 Created Time  :2015-8-3 14:17:24
  9 File Name     :POJ_3414_BFS.cpp
 10 **************************************************/
 11 
 12 #include <cstdio>
 13 #include <algorithm>
 14 #include <iostream>
 15 #include <sstream>
 16 #include <cstring>
 17 #include <cmath>
 18 #include <string>
 19 #include <vector>
 20 #include <queue>
 21 #include <deque>
 22 #include <stack>
 23 #include <list>
 24 #include <map>
 25 #include <set>
 26 #include <bitset>
 27 #include <cstdlib>
 28 #include <ctime>
 29 using namespace std;
 30 
 31 #define lson l, mid, rt << 1
 32 #define rson mid + 1, r, rt << 1 | 1
 33 typedef long long ll;
 34 const int MAXN = 1e4 + 10;
 35 const int INF = 0x3f3f3f3f;
 36 const int MOD = 1e9 + 7;
 37 struct Point    {
 38     int a, b, step;
 39     int op[MAXN];
 40 };
 41 bool vis[110][110];
 42 int A, B, C;
 43 int ans;
 44 
 45 void BFS(void)  {
 46     memset (vis, false, sizeof (vis));
 47     queue<Point> Q; Q.push ((Point) {0, 0, 0});
 48     while (!Q.empty ()) {
 49         Point p = Q.front ();   Q.pop ();
 50         if (p.a == C || p.b == C)   {
 51             printf ("%d\n", p.step);
 52             for (int i=1; i<=p.step; ++i)   {
 53                 if (p.op[i] == 1)         puts ("FILL(1)");
 54                 else if (p.op[i] == 2)    puts ("FILL(2)");
 55                 else if (p.op[i] == 3)    puts ("DROP(1)");
 56                 else if (p.op[i] == 4)    puts ("DROP(2)");
 57                 else if (p.op[i] == 5)    puts ("POUR(1,2)");
 58                 else if (p.op[i] == 6)    puts ("POUR(2,1)");
 59             }
 60             return ;
 61         }
 62         Point tmp;
 63         if (p.a < A && !vis[A][p.b])    {
 64             vis[A][p.b] = true;
 65             tmp = p;    tmp.a = A;  tmp.op[++tmp.step] = 1;     //FILL1
 66             Q.push (tmp);
 67         }
 68         if (p.b < B && !vis[p.a][B])    {
 69             vis[p.a][B] = true;
 70             tmp = p;    tmp.b = B;  tmp.op[++tmp.step] = 2;     //FILL2
 71             Q.push (tmp);
 72         }
 73         if (p.a > 0 && !vis[0][p.b])    {
 74             vis[0][p.b] = true;
 75             tmp = p;    tmp.a = 0;  tmp.op[++tmp.step] = 3;     //DROP1
 76             Q.push (tmp);
 77         }
 78         if (p.b > 0 && !vis[p.a][0])    {
 79             vis[p.a][0] = true;
 80             tmp = p;    tmp.b = 0;  tmp.op[++tmp.step] = 4;     //DROP2
 81             Q.push (tmp);
 82         }
 83         if (p.a > 0 && p.b < B) {
 84             int t = min (p.a, B - p.b);
 85             if (!vis[p.a-t][p.b+t]) {
 86                 vis[p.a-t][p.b+t] = true;                       //POUR1->2
 87                 tmp = p;    tmp.a -= t; tmp.b += t; tmp.op[++tmp.step] = 5;
 88                 Q.push (tmp);
 89             }
 90         }
 91         if (p.b > 0 && p.a < A) {
 92             int t = min (p.b, A - p.a);
 93             if (!vis[p.a+t][p.b-t]) {
 94                 vis[p.a+t][p.b-t] = true;                       //POUR2->1
 95                 tmp = p;    tmp.a += t; tmp.b -= t; tmp.op[++tmp.step] = 6;
 96                 Q.push (tmp);
 97             }
 98         }
 99     }
100 
101     puts ("impossible");
102 }
103 
104 int main(void)    {       //POJ 3414 Pots
105     while (scanf ("%d%d%d", &A, &B, &C) == 3) {
106         BFS ();
107     }
108 
109     return 0;
110 }

 

posted @ 2015-08-03 20:54  Running_Time  阅读(174)  评论(0)  编辑  收藏  举报