POJ3414Pots【BFS】

大意：有两个杯子  三种操作

You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:

  1. FILL(i)        fill the pot i (1 ≤ i ≤ 2) from the tap;
  2. DROP(i)      empty the pot i to the drain;
  3. POUR(i,j)    pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j)

问要达到指定的容量需要操作的最少次数

 

分析：

BFS扩展每种情况然后再输出即可

代码：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <queue>
  5 using namespace std;
  6 
  7 const int maxn = 1000;
  8 
  9 struct Node {
 10     int x, y;
 11     int caozuo;
 12     int a, b;
 13     int fa, now;
 14     int num;
 15 }node[maxn + 10];
 16 
 17 queue<Node> q;
 18 int vis[105][105];
 19 
 20 void dfs(Node n0) {
 21     if(n0.caozuo == 0) {
 22         return ;
 23     }
 24     dfs(node[n0.fa + 1]);
 25     if(n0.caozuo == 1) {
 26         printf("FILL(%d)\n", n0.a + n0.b);
 27     } else if(n0.caozuo == 2) {
 28         printf("DROP(%d)\n", n0.a + n0.b);
 29     } else {
 30         printf("POUR(%d,%d)\n", n0.a, n0.b);
 31     }
 32 }
 33 
 34 int main() {
 35     int A, B, C;
 36     while(EOF != scanf("%d %d %d",&A, &B, &C) ) {
 37         memset(vis, 0, sizeof(vis));
 38         int cnt = 0;
 39         node[cnt++] = (Node) { 0, 0, 0, 0, 0, 0, 0, 0 };
 40         q.push(node[0]); vis[0][0] = 1;
 41         bool flag = false;
 42         int x = 0;
 43         Node ans;
 44         cnt++;
 45         while(!q.empty()) {
 46             if(x++ > maxn) break;
 47 
 48             Node n0 = q.front(); q.pop();
 49 //            printf("now = %d   x = %d   y = %d   fa = %d  num = %d  caozuo = %d\n", n0.now, n0.x, n0.y, n0.fa, n0.num, n0.caozuo);
 50 
 51             if(n0.x == C || n0.y == C) {
 52                 flag = true;
 53                 ans = n0;
 54                 break;
 55             }
 56 
 57             if(!vis[A][n0.y]) {
 58                 node[cnt++] = (Node) { A, n0.y, 1, 1, 0, n0.now, cnt - 1, n0.num + 1};
 59                 q.push(node[cnt - 1]);
 60                 vis[A][n0.y] = 1;
 61             }
 62             
 63             if(!vis[n0.x][B]) {
 64                 node[cnt++] = (Node) { n0.x, B, 1, 1, 1, n0.now, cnt - 1,n0.num + 1 };
 65                 q.push(node[cnt - 1]);
 66                 vis[n0.x][B] = 1;
 67             }
 68 
 69             if(!vis[0][n0.y]) {
 70                 node[cnt++] = (Node) { 0, n0.y, 2, 1, 0, n0.now, cnt - 1,n0.num + 1 };
 71                 q.push(node[cnt - 1]);
 72                 vis[0][n0.y] = 1;
 73             }
 74             
 75             if(!vis[n0.x][0]) {
 76                 node[cnt++] = (Node) { n0.x, 0, 2, 1, 1, n0.now, cnt - 1,n0.num + 1 };
 77                 q.push(node[cnt - 1]);
 78                 vis[n0.x][0] = 1;
 79             }
 80 
 81             if(n0.x + n0.y <= B) {
 82                 if(!vis[0][n0.x + n0.y]) {
 83                     node[cnt++] = (Node) { 0, n0.x + n0.y, 3, 1, 2, n0.now, cnt - 1, n0.num + 1};
 84                     q.push(node[cnt -1]);
 85                     vis[0][n0.x + n0.y] = 1;
 86                 }
 87             } else {
 88                 if(!vis[n0.x + n0.y - B][B]) {
 89                     node[cnt++] = (Node) { n0.x + n0.y - B, B, 3, 1, 2, n0.now, cnt - 1,n0.num + 1 };
 90                     q.push(node[cnt - 1]);
 91                     vis[n0.x + n0.y - B][B] = 1;
 92                 }
 93             }
 94 
 95             if(n0.x + n0.y <= A) {
 96                 if(!vis[n0.x + n0.y][0]) {
 97                     node[cnt++] = (Node) { n0.x + n0.y, 0, 3, 2, 1, n0.now, cnt - 1,n0.num + 1};
 98                     q.push(node[cnt -1]);
 99                     vis[n0.x + n0.y][0] = 1;
100                 }
101             } else {
102                 if(!vis[A][n0.x + n0.y - A]) {
103                     node[cnt++] = (Node) { A, n0.x + n0.y - A, 3, 2, 1, n0.now, cnt - 1,n0.num + 1};
104                     q.push(node[cnt - 1]);
105                     vis[A][n0.x + n0.y - A] = 1;
106                 }
107             }
108         }
109         if(!flag) {
110             puts("impossible");
111             continue;
112         }
113         printf("%d\n", ans.num);
114         Node n0 = ans;
115         dfs(n0); 
116     }
117     return 0;
118 }
View Code

 

posted @ 2014-11-01 08:53  悠悠我心。  阅读(265)  评论(0)  编辑  收藏  举报