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 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.
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).
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’.
3 5 4
6 FILL(2) POUR(2,1) DROP(1) POUR(2,1) FILL(2) POUR(2,1)
bfs的搜索永遠優(yōu)先最小操作數(shù)的，所以再用dp[][]記錄防止重復即可在o(A*B)找到答案。
為了方便回溯路徑，用一個back[][]記錄前驅(qū)即可
//Accepted 760kB 0ms#include<stdio.h>#include<string.h>#include<algorithm>#include<queue>#define MAX_N 102using namespace std;struct node{int last_A,last_B,op;};int dp[MAX_N][MAX_N];node back[MAX_N][MAX_N];int A,B,goal;char str[7][10]={"nothing","FILL(1)","FILL(2)","DROP(2)","DROP(1)","POUR(1,2)","POUR(2,1)"};void print(int n_A,int n_B){ node e=back[n_A][n_B]; if(e.last_A||e.last_B) print(e.last_A,e.last_B); puts(str[e.op]);}void bfs(){ queue<pair<int,int> >que;//first-> A second-> B memset(dp,-1,sizeof(dp)); que.push(make_pair(0,0)); dp[0][0]=0;//dp record min steps from start while(!que.empty()){ int n_A=que.front().first,n_B=que.front().second;que.pop(); int s=dp[n_A][n_B]; if(n_A==goal||n_B==goal){//back printf("%d/n",s); print(n_A,n_B); return ; } //操作一步有六種情況 if(dp[A][n_B]==-1){//FILL(1) dp[A][n_B]=s+1; back[A][n_B]={n_A,n_B,1}; que.push(make_pair(A,n_B)); } if(dp[n_A][B]==-1){//FILL(2) dp[n_A][B]=s+1; back[n_A][B]={n_A,n_B,2}; que.push(make_pair(n_A,B)); } if(dp[n_A][0]==-1){//DROP(2) dp[n_A][0]=s+1; back[n_A][0]={n_A,n_B,3}; que.push(make_pair(n_A,0)); } if(dp[0][n_B]==-1){//DROP(1) dp[0][n_B]=s+1; back[0][n_B]={n_A,n_B,4}; que.push(make_pair(0,n_B)); } int tmpb=n_A+n_B>B?B:n_A+n_B; int tmpa=n_A+n_B-tmpb; if(dp[tmpa][tmpb]==-1){//POUR(1,2) dp[tmpa][tmpb]=s+1; back[tmpa][tmpb]={n_A,n_B,5}; que.push(make_pair(tmpa,tmpb)); } tmpa=n_A+n_B>A?A:n_A+n_B; tmpb=n_A+n_B-tmpa; if(dp[tmpa][tmpb]==-1){//POUR(2,1) dp[tmpa][tmpb]=s+1; back[tmpa][tmpb]={n_A,n_B,6}; que.push(make_pair(tmpa,tmpb)); } } puts("impossible");}int main(){ while(~scanf("%d%d%d",&A,&B,&goal)){ bfs(); } return 0;}新聞熱點
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