简单搜索：poj3414

Posted 2020-09-18

 Pots

题意：两个容器，容量A，B，进行倒水操作，求最少的操作次数使得某一容器水量为C。

思路：一个多月前写的时候是没思路的，但是随着搜索算法的学习，慢慢懂得了一些 状态 的问题。这里，倒水就有很多的状态比如（FILL(A),B）(A, FILL(B))等。意识到这个之后，因为要求的是最少的操作次数，所以用bfs，注意剪枝。代码如下：

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <queue>
  4 #include <vector>
  5 using namespace std;
  6 
  7 int A, B, C;
  8 int hashed[11000];
  9 
 10 struct Node {
 11     int la, lb;
 12     int step;
 13     vector<int>path;
 14 };
 15 
 16 const char *Path[] = {"HAHA\0", "DROP(1)\0", "DROP(2)\0", "FILL(1)\0", "FILL(2)\0",
 17                       "POUR(1,2)\0", "POUR(2,1)\0"};
 18 
 19 bool is_ok(Node node)
 20 {
 21     int idx = node.la * 100 + node.lb;
 22     if (hashed[idx] == 1)
 23         return false;
 24     hashed[idx] = 1;
 25     return true;
 26 }
 27 
 28 int bfs()
 29 {
 30     queue<Node>Q;
 31     Node node;
 32     node.la = 0, node.lb = 0, node.path.clear(), node.step = 0;
 33     Q.push(node);
 34 
 35     while (!Q.empty()) {
 36         Node now = Q.front(); Q.pop();
 37         if (now.la == C || now.lb == C) {
 38             printf("%d\n", now.step);
 39             for (vector<int>::iterator it = now.path.begin(); it != now.path.end(); it++) {
 40                 printf("%s\n", Path[*it]);
 41             }
 42             return 1;
 43         }
 44         for (int i = 1; i <= 6; i++) {
 45             if (i == 1) {
 46                 Node next = now;
 47                 next.la = 0, next.lb = now.lb, next.path.push_back(i), next.step = now.step + 1;
 48                 if (next.lb != 0 && is_ok(next))
 49                     Q.push(next);
 50             }
 51             if (i == 2) {
 52                 Node next = now;
 53                 next.la = now.la, next.lb = 0, next.path.push_back(i), next.step = now.step + 1;
 54                 if (next.la != 0 && is_ok(next))
 55                     Q.push(next);
 56             }
 57             if (i == 3) {
 58                 Node next = now;
 59                 next.la = A, next.lb = now.lb, next.path.push_back(i), next.step = now.step + 1;
 60                 if (is_ok(next))
 61                 Q.push(next);
 62             }
 63             if (i == 4) {
 64                 Node next = now;
 65                 next.la = now.la, next.lb = B, next.path.push_back(i), next.step = now.step + 1;
 66                 if (is_ok(next))
 67                 Q.push(next);
 68             }
 69             if (i == 5) {
 70                 Node next = now;
 71                 int su = now.la + now.lb;
 72                 if (su >= B) {
 73                     next.la = su - B, next.lb = B, next.path.push_back(i), next.step = now.step + 1;
 74                     if (is_ok(next))
 75                     Q.push(next);
 76                 } else {
 77                     next.la = 0, next.lb = su, next.path.push_back(i), next.step = now.step + 1;
 78                     if (is_ok(next))
 79                     Q.push(next);
 80                 }
 81             }
 82             if (i == 6) {
 83                 Node next = now;
 84                 int su = now.la + now.lb;
 85                 if (su >= A) {
 86                     next.la = A, next.lb = su - A, next.path.push_back(i), next.step = now.step + 1;
 87                     if (is_ok(next))
 88                     Q.push(next);
 89                 } else {
 90                     next.la = su, next.lb = 0, next.path.push_back(i), next.step = now.step + 1;
 91                     if (is_ok(next))
 92                     Q.push(next);
 93                 }
 94             }
 95         }
 96     }
 97     return -1;
 98 }
 99 
100 int main()
101 {
102     scanf("%d%d%d", &A, &B, &C);
103     int ans = bfs();
104     if (ans == -1 ) {
105         printf("impossible\n");
106     }
107     return 0;
108 }

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