Pots

Time Limit: 1000ms?? Memory limit: 65536K??有疑問？點這里^_^

題目描寫敘述

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.

輸入

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).

輸出

The first line of the output must contain the length of the sequence of operations K. ?If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.

演示樣例輸入

3 5 4

演示樣例輸出

6

提示

FILL(2) POUR(2,1) DROP(1) POUR(2,1) FILL(2) POUR(2,1)
簡單搜索題，對于兩個空瓶子，容積分別為A B 有6種操作 把A（或B）清空，把A（或B）裝滿，把A倒入B，把B倒入A 。相應這6種操作，有6種狀態。典型的bfs搜索。不多了，僅僅是這題明明說的是單組輸入結果答案卻要多組輸入才對。白白貢獻5個WA。 #include <cstdio> #include <iostream> #include <cstring> #include <cstdlib> #include <queue> using namespace std; int m,n,c; typedef struct node {int v1,v2,op; }; bool vis[999][999]; void bfs() {node t={0,0,0};queue <node> Q;Q.push(t);vis[0][0]=1;while(!Q.empty()){node f=Q.front();Q.pop();if(f.v1==c||f.v2==c){cout<<f.op<<endl;return ;}if(f.v1!=m){t.v1=m;t.op=f.op+1;t.v2=f.v2;if(!vis[t.v1][t.v2]){vis[t.v1][t.v2]=1;Q.push(t);}}if(f.v2!=n){t.v2=n;t.op=f.op+1;t.v1=f.v1;if(!vis[t.v1][t.v2]){vis[t.v1][t.v2]=1;Q.push(t);}}if(f.v1!=0){t.v1=0;t.v2=f.v2;t.op=f.op+1;if(!vis[t.v1][t.v2]){vis[t.v1][t.v2]=1;Q.push(t);}}if(f.v2!=0){t.v2=0;t.v1=f.v1;t.op=f.op+1;if(!vis[t.v1][t.v2]){vis[t.v1][t.v2]=1;Q.push(t);}}if(f.v2!=0&&f.v1!=m){t.v2=f.v2-(m-f.v1);if(t.v2<0) t.v2=0;t.v1=f.v1+f.v2; if(t.v1>m) t.v1=m;t.op=f.op+1;if(!vis[t.v1][t.v2]){vis[t.v1][t.v2]=1;Q.push(t);}}if(f.v1!=0&&f.v2!=n){t.v1=f.v1-(n-f.v2);if(t.v1<0) t.v1=0;t.v2=f.v2+f.v1; if(t.v2>n) t.v2=n;t.op=f.op+1;if(!vis[t.v1][t.v2]){vis[t.v1][t.v2]=1;Q.push(t);}}}puts("impossible"); } int main() {while(cin>>m>>n>>c){memset(vis,0,sizeof(vis));bfs();}return 0; }

轉載于:https://www.cnblogs.com/jzssuanfa/p/7026035.html

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