POJ 3414 Pots

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题目链接：http://poj.org/problem?id=3414

题目描述：

给定三个数A、B、C，表示现有容积为A、B的两个容器，有六种操作，初始状态容器为空，问是否能够经过操作使得中间至少一个容器有C个单位液体。

因为A、B、C的大小均小于100，所以无论能否组成，尝试的最多A、B组合的情况是 100 * 100 ，求最小的步数，直接BFS即可，可以用哈希判重。找到过，往回输出步骤。

<span style="font-size:18px;">#include<cstdio>#include<iostream>#include<string>#include<cstring>#include<queue>using namespace std ;const int MAXM = 1e5 ;const int MAXN = 1e4 + 10 ;int A, B, C ;bool used[MAXM] ;int state[MAXN] ; //FILLa, b; DROPa, b; POURa,b, b,a(1 ~ 6)int res[MAXN] ;int pos, cur ;string pri[7] = {"", "FILL(1)", "FILL(2)", "DROP(1)", "DROP(2)", "POUR(1,2)", "POUR(2,1)"} ; //struct node{int num, aa, bb, point ;}rec[MAXN] ;//void setValue(int a1, int a2, int a3, int a4){rec[cur].num = a1 ;rec[cur].aa = a2 ;rec[cur].bb = a3 ;rec[cur].point = a4 ;}//void cal(){int p ;pos = -1, cur = 0 ;memset(used, false, sizeof(used)) ;setValue(0, 0, 0, 0) ;used[0] = true ;pos++ ;p = -1 ;while(  pos != -1 ){p++ ;int vp = p ;int vv = rec[p].num ;int va = rec[p].aa ;int vb = rec[p].bb ;pos-- ;//if( va == C || vb == C ){cout << vv << endl ;int tt = 0 ;res[tt++] = state[p] ;while( rec[p].point != 0 ){res[tt++] = state[rec[p].point] ;p = rec[p].point ;}for( int i = tt-1; i >= 0; i-- ){cout << pri[res[i]] << endl ;}return ;}//int tmp1, tmp2 ;//FILLtmp1 = 105 * A + vb ;if( !used[tmp1] ){used[tmp1] = true ;cur++, pos++ ;state[cur] = 1 ;setValue(vv+1, A, vb, vp) ;}tmp2 = 105 * va + B ;if( !used[tmp2] ){used[tmp2] = true ;cur++, pos++ ;state[cur] = 2 ;setValue(vv+1, va, B, vp) ;}//DROPtmp1 = vb ;if( !used[tmp1] ){used[tmp1] = true ;cur++, pos++ ;state[cur] = 3 ;setValue(vv+1, 0, vb, vp) ;}tmp2 = 105 * va ;if( !used[tmp2] ){used[tmp2] = true ;cur++, pos++ ;state[cur] = 4 ;setValue(vv+1, va, 0, vp) ;}//POURif( vb + va < B ){tmp1 = vb + va ;if( !used[tmp1] ){used[tmp1] = true ;cur++, pos++ ;state[cur] = 5 ;setValue(vv+1, 0, va+vb, vp) ;}}else{tmp1 = 105 * (va+vb-B) + B ; if( !used[tmp1] ){used[tmp1] = true ;cur++, pos++ ;state[cur] = 5 ;setValue(vv+1, va+vb-B, B, vp) ;}}//if( vb + va < A ){tmp2 =  105 * (va+vb) ;if( !used[tmp2] ){used[tmp2] = true ;cur++, pos++ ;state[cur] = 6 ;setValue(vv+1, va+vb, 0, vp) ;}}else{tmp2 = 105 * A + (va+vb-A) ; if( !used[tmp2] ){used[tmp2] = true ;cur++, pos++ ;state[cur] = 6 ;setValue(vv+1, A, va+vb-A, vp) ;}}}cout << "impossible" << endl ;}//int main(){/////freopen("1234.txt", "r", stdin) ;while( cin >> A >> B >> C ){if( A == C || B == C ){cout << 1 << endl ;if( A == C ){cout << "FILL(1)" << endl ;}else{cout << "FILL(2)" << endl ;}}else{cal() ;}}return 0 ;}</span>

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