20130421.4 倒水问题(BFS)

【编程题】
给你两个烧杯，其容积分别是A升和B升。允许进行下列操作：
1．FILL(i)            从水管接水将烧杯i灌满（(1 ≤ i ≤ 2）
2．DROP(i)        将第i个烧杯倒空
3．POUR(i,j)         将烧杯i的水倒进烧杯j；并且倒完后要么烧杯j是满的（烧杯i中也许有剩水），要么烧杯i是空的（此时烧杯i的水应该全部倒进烧杯j）。
写一个程序寻找最少的操作序列，使得这两个烧杯中的某一个烧杯中只剩C升水。

输入

仅仅1行，是数字A,B,C。它们都是整数，范围1到100并且C≤max(A,B)。

输出

第1行是操作序列的长度K。接下来的K行中，每行描述一次操作。如果存在多个最小长度序列，可以输出它们中间的任何一个。如果目标不可达，输出impossible。

样例输入

3 5 4

样例输出

6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)

(简单的宽度优先搜索,不过要先判断最多多少种情况)

#include<iostream>#include<stdlib.h>#include<stack>using namespace std;stack<int> s;typedef struct{int a;int b;int fa;int op;}ST;char operation[6][20] = {"FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)"};int vis[100][100];ST arr[10000];int main(){ST now;int A,B,C;int prior,next,pos;cin>>A>>B>>C;arr[0].a = 0;arr[0].b = 0;arr[0].fa = -1;prior = 0; next = 1;while(prior < next){now = arr[prior];if(now.a == C || now.b == C)break;if(now.a != A){arr[next].a = A;arr[next].b = now.b;arr[next].fa = prior;arr[next].op = 0;if(!vis[arr[next].a][arr[next].b]){vis[arr[next].a][arr[next].b] = 1;next++;}}if(now.b != B){arr[next].b = B;arr[next].a = now.a;arr[next].fa = prior;arr[next].op = 1;if(!vis[arr[next].a][arr[next].b]){vis[arr[next].a][arr[next].b] = 1;next++;}}if(now.a != 0){arr[next].a = 0;arr[next].b = now.b;arr[next].fa = prior;arr[next].op = 2;if(!vis[arr[next].a][arr[next].b]){vis[arr[next].a][arr[next].b] = 1;next++;}}if(now.b != 0){arr[next].b = 0;arr[next].a = now.a;arr[next].fa = prior;arr[next].op = 3;if(!vis[arr[next].a][arr[next].b]){vis[arr[next].a][arr[next].b] = 1;next++;}}if(now.a != 0 && now.b != B){if(now.a + now.b <= B){arr[next].a = 0;arr[next].b = now.a + now.b;}else{arr[next].a = now.a - (B-now.b);arr[next].b = B;}arr[next].fa = prior;arr[next].op = 4;if(!vis[arr[next].a][arr[next].b]){vis[arr[next].a][arr[next].b] = 1;next++;}}if(now.b != 0 && now.a != A){if(now.a + now.b <= A){arr[next].b = 0;arr[next].a = now.a + now.b;}else{arr[next].b = now.b - (A-now.a);arr[next].a = A;}arr[next].fa = prior;arr[next].op = 5;if(!vis[arr[next].a][arr[next].b]){vis[arr[next].a][arr[next].b] = 1;next++;}}prior++;}if(prior >= next) cout<<"impossible"<<endl;else{pos = prior;while(arr[pos].fa != -1){s.push(arr[pos].op);pos = arr[pos].fa;}while(!s.empty()){cout<<operation[s.top()]<<endl;s.pop();}}//system("pause");return 0;}