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| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 11703 | Accepted: 4955 | Special Judge | ||
Description
You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:
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)
好爽,一遍ac
/************************************************************************* > File Name: code/2015summer/searching/H.cpp > Author: 111qqz > Email: rkz2013@126.com > Created Time: 2015年07月27日 星期一 09时11分28秒 ************************************************************************/ #include<iostream> #include<iomanip> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<string> #include<map> #include<set> #include<queue> #include<vector> #include<stack> #define y0 abc111qqz #define y1 hust111qqz #define yn hez111qqz #define j1 cute111qqz #define tm crazy111qqz #define lr dying111qqz using namespace std; #define REP(i, n) for (int i=0;i<int(n);++i) typedef long long LL; typedef unsigned long long ULL; const int N=1E2+5; int A,B,C; int d[N][N]; bool flag; struct node { int d,opt,par,prea,preb; }q[N][N]; void print(int x,int y) { // cout<<"x:"<<x<<"y:"<<y<<endl; if (q[x][y].prea!=-1&&q[x][y].preb!=-1) { // cout<<"who is 111qqz"<<endl; print(q[x][y].prea,q[x][y].preb); if (q[x][y].opt==1){ printf("FILL(%d)\n",q[x][y].par); } if (q[x][y].opt==2) { printf("DROP(%d)\n",q[x][y].par); } if (q[x][y].opt==3) { printf("POUR(%d,%d)\n",q[x][y].par,3-q[x][y].par); } } } void bfs() { memset(q,-1,sizeof(q)); queue<int>a; queue<int>b; a.push(0); b.push(0); q[0][0].d=0; while (!a.empty()&&!b.empty()) { int av = a.front();a.pop(); int bv = b.front();b.pop(); // cout<<"av:"<<av<<"bv:"<<bv<<endl; if (av==C||bv==C) { flag = true; //cout<<"yeah~~~~~~~~~~~~~~~"<<endl; cout<<q[av][bv].d<<endl; // cout<<"prea:"<<q[av][bv].prea<<"preb:"<<q[av][bv].preb<<endl; print(av,bv); return; } if (av<A&&q[A][bv].d==-1) { q[A][bv].d=q[av][bv].d+1; q[A][bv].opt=1; q[A][bv].par=1; q[A][bv].prea=av; q[A][bv].preb=bv; a.push(A); b.push(bv); } if (av>0&&q[0][bv].d==-1) { q[0][bv].d=q[av][bv].d+1; q[0][bv].opt=2; q[0][bv].par=1; q[0][bv].prea=av; q[0][bv].preb=bv; a.push(0); b.push(bv); } if (bv<B&&q[av][B].d==-1) { q[av][B].d=q[av][bv].d+1; q[av][B].opt=1; q[av][B].par=2; q[av][B].prea = av; q[av][B].preb = bv; a.push(av); b.push(B); } if (bv>0&&q[av][0].d==-1) { q[av][0].d=q[av][bv].d+1; q[av][0].opt=2; q[av][0].par=2; q[av][0].prea=av; q[av][0].preb=bv; a.push(av); b.push(0); } if (av+bv<=B&&q[0][av+bv].d==-1) { q[0][av+bv].d=q[av][bv].d+1; q[0][av+bv].opt=3; q[0][av+bv].par=1; q[0][av+bv].prea=av; q[0][av+bv].preb=bv; a.push(0); b.push(av+bv); } if (av+bv>B&&q[av-(B-bv)][B].d==-1) //把1往2里倒入的两种情况 { int tmp = av-(B-bv); q[tmp][B].d=q[av][bv].d+1; q[tmp][B].opt=3; q[tmp][B].par=1; q[tmp][B].prea=av; q[tmp][B].preb=bv; a.push(tmp); b.push(B); } if (bv+av<=A&&q[av+bv][0].d==-1) { q[av+bv][0].d=q[av][bv].d+1; q[av+bv][0].opt=3; q[av+bv][0].par=2; q[av+bv][0].prea=av; q[av+bv][0].preb=bv; a.push(av+bv); b.push(0); } if (bv+av>A&&q[A][bv-(A-av)].d==-1) { int tmp = bv-(A-av); q[A][tmp].d=q[av][bv].d+1; q[A][tmp].opt=3; q[A][tmp].par=2; q[A][tmp].prea=av; q[A][tmp].preb=bv; a.push(A); b.push(tmp); } } } int main() { flag = false; cin>>A>>B>>C; bfs(); if (!flag) { cout<<"impossible"<<endl; } return 0; }
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原文地址：http://www.cnblogs.com/111qqz/p/4680562.html
