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poj3414--Pots(bfs,记录路径)
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 10149 | Accepted: 4275 | Special Judge | ||
Description
You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:
- FILL(i) fill the pot i (1 ≤ i ≤ 2) from the tap;
- DROP(i) empty the pot i to the drain;
- POUR(i,j) pour from pot i to pot j; after this operation either the potj is full (and there may be some water left in the pot i), or the poti 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 exactlyC liters of water in one of the pots.
Input
On the first and only line are the numbers A, B, andC. 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 operationsK. 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
6FILL(2)POUR(2,1)DROP(1)POUR(2,1)FILL(2)POUR(2,1)
有a，b两个瓶，得到体积为c的液体，六种操作
装满1，装满2，倒掉1，倒掉2，由1倒到2，有2倒到1.
记录一下路径，也就是有谁到达的当前状态
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct node{ int a , b , pre ; 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(1,2)","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;}
poj3414--Pots(bfs,记录路径)