Pots
You are given two pots, having the volume of A and B liters respectively. The following operations can be performed: FILL(i): fill the poti (1 ≤ i ≤ 2) from the tap; DROP(i): empty the poti to the drain; POUR(i,j): pour from poti 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).
Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.
Input
On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).
Output
The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.
Sample Input
3 5 4
Sample Output
6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)
Source
Northeastern Europe 2002, Western Subregion
题目类型：BFS+记录路径
算法分析：注意在记录路径的时候要分好类
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#pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7FFFFFFF; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e5 + 6666; int A, B, C, vis[166][166]; struct node { int a, b, cnt; node (int aa, int bb, int ccnt) : a (aa), b (bb), cnt (ccnt) {} node () {} }; struct point { int a, b, flag; }; point par[166][166], out[666]; int pa, pb; int bfs () { memset (vis, false, sizeof (vis)); for (int i = 0; i < 166; i++) for (int j = 0; j < 166; j++) par[i][j].a = par[i][j].b = par[i][j].flag = -1; vis[0][0] = true; queue <node> qu; qu.push (node (0, 0, 0)); while (!qu.empty ()) { node tt = qu.front (); qu.pop (); if (tt.a == C || tt.b == C) { pa = tt.a, pb = tt.b; return tt.cnt; } int ta = tt.a, tb = B; if (!vis[ta][tb]) { vis[ta][tb] = true; qu.push (node (ta, tb, tt.cnt + 1)); par[ta][tb].a = tt.a; par[ta][tb].b = tt.b; par[ta][tb].flag = 1; } ta = A, tb = tt.b; if (!vis[ta][tb]) { vis[ta][tb] = true; qu.push (node (ta, tb, tt.cnt + 1)); par[ta][tb].a = tt.a; par[ta][tb].b = tt.b; par[ta][tb].flag = 2; } ta = 0, tb = tt.b; if (!vis[ta][tb]) { vis[ta][tb] = true; qu.push (node (ta, tb, tt.cnt + 1)); par[ta][tb].a = tt.a; par[ta][tb].b = tt.b; par[ta][tb].flag = 3; } ta = tt.a, tb = 0; if (!vis[ta][tb]) { vis[ta][tb] = true; qu.push (node (ta, tb, tt.cnt + 1)); par[ta][tb].a = tt.a; par[ta][tb].b = tt.b; par[ta][tb].flag = 4; } int tk = min (tt.a, B - tt.b); ta = tt.a - tk, tb = tt.b + tk; if (!vis[ta][tb]) { vis[ta][tb] = true; qu.push (node (ta, tb, tt.cnt + 1)); par[ta][tb].a = tt.a; par[ta][tb].b = tt.b; par[ta][tb].flag = 5; } tk = min (tt.b, A - tt.a); ta = tt.a + tk, tb = tt.b - tk; if (!vis[ta][tb]) { vis[ta][tb] = true; qu.push (node (ta, tb, tt.cnt + 1)); par[ta][tb].a = tt.a; par[ta][tb].b = tt.b; par[ta][tb].flag = 6; } } return -1; } int main() { // CFF; while (scanf ("%d %d %d", &A, &B, &C) != EOF) { int tt = bfs (); if (tt == -1) puts ("impossible"); else { printf ("%d\n", tt); int len = 0; do { out[len].a = par[pa][pb].a; out[len].b = par[pa][pb].b; out[len++].flag = par[pa][pb].flag; int tta = par[pa][pb].a, ttb = par[pa][pb].b; pa = tta, pb = ttb; } while (pa != -1 && pb != -1); for (int i = len - 1; i >= 0; i--) { if (out[i].flag == 1) puts ("FILL(2)"); if (out[i].flag == 2) puts ("FILL(1)"); if (out[i].flag == 3) puts ("DROP(1)"); if (out[i].flag == 4) puts ("DROP(2)"); if (out[i].flag == 5) puts ("POUR(1,2)"); if (out[i].flag == 6) puts ("POUR(2,1)"); } } } return 0; } |
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