Poj3414--pots (BFS, record Path)

Pots

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10149		Accepted: 4275		Special Judge

Description

You are given the pots of 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 potJ was full (and there may be some water left in the pot I), or the pot i was empty (and all its contents has been moved to the pot J).

Write a program to find the shortest possible sequence of these operations that would yield exactlyC liters of W Ater in one of the pots.

Input

On the first and only line is the numbers A, B, andC. These is all integers in the range from 1 to + C≤max (A,B).

Output

The first line of the output must contain the length of the sequence of operationsK. The following K lines must each describe one operation. If There is several sequences of minimal length, output any one of them. If the desired result can ' t is achieved, the first and only line of the file must contain the word 'impossible'.

Sample Input

3 5 4

Sample Output

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

There is a A, b two bottles, get the volume of C liquid, six kinds of operation

Fill 1. Fill 2, pour 1. Pour 2, pour from 1 to 2, 2 pour to 1.

The

records the path. That is, who has reached the current state

#include <cstdio> #include <cstring> #include <algorithm>using namespace std;struct node{int A, B, p    Re; Char s[20];} P[1000000], x; int flag[20000], low, top, K; char str[6][20] = {"Fill (1)", "Fill (2)", "Drop (1)", "Drop (2)", "pour", "    Pour (2,1) "};int pre[1000000"; int main () {int A, B, C, Sum, I, J;    k = 0;    memset (flag,0,sizeof (flag));    scanf ("%d%d%d", &a, &b, &c);    FLAG[A*100+B] = 1;    Low = top = 0; P[TOP].A = 0;    p[top].b = 0;    P[top++].pre =-1;        while (Low < top) {if (p[low].a = = c | | p[low].b = = c) break;            if (P[low].a < a) {p[top].a = A; p[top].b = p[low].b;                if (!flag[p[top].a*100+p[top].b]) {flag[p[top].a*100+p[top].b] = 1;                strcpy (P[top].s,str[0]);            P[top++].pre = low; }} if (P[low].b < b) {p[top].a = p[low].a; p[top].b = b;           if (!flag[p[top].a*100+p[top].b]) {flag[p[top].a*100+p[top].b] = 1;                strcpy (p[top].s,str[1]);            P[top++].pre = low;            }} if (P[low].a > 0) {p[top].a = 0; p[top].b = p[low].b;                if (!flag[p[top].a*100+p[top].b]) {flag[p[top].a*100+p[top].b] = 1;                strcpy (p[top].s,str[2]);            P[top++].pre = low;            }} if (p[low].b > 0) {p[top].a = p[low].a; p[top].b = 0;                if (!flag[p[top].a*100+p[top].b]) {flag[p[top].a*100+p[top].b] = 1;                strcpy (P[top].s,str[3]);            P[top++].pre = low;            }} if (P[low].a > 0 && p[low].b < b) {sum = p[low].a + p[low].b;      if (sum <= b) {p[top].a = 0; p[top].b = sum;      } else {p[top].a = sum-b; p[top].b = b;                } if (!flag[p[top].a*100+p[top].b]) {flag[p[top].a*100+p[top].b] = 1;                strcpy (P[top].s,str[4]);            P[top++].pre = low;            }} if (P[low].a < a && p[low].b > 0) {sum = p[low].a + p[low].b;            if (sum <= a) {p[top].a = sum; p[top].b = 0;            } else {p[top].a = A; p[top].b = sum-a;                } if (!flag[p[top].a*100+p[top].b]) {flag[p[top].a*100+p[top].b] = 1;                strcpy (P[top].s,str[5]);            P[top++].pre = low;    }} low++;    } if (low = = top) printf ("impossible\n");        else {j = 0;   for (i = low; I! = 0; i = p[i].pre) {pre[j++] = i;     } printf ("%d\n", j);    for (J-= 1; J >= 0; j--) printf ("%s\n", p[Pre[j]].s); } return 0;}

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Poj3414--pots (BFS, record Path)