《算法笔记》8.2小节——搜索专题-＞广度优先搜索（BFS）问题 A: Jugs
题目描述
In the movie "Die Hard 3", Bruce Willis and Samuel L. Jackson were confronted with the following puzzle. They were given a 3-gallon jug and a 5-gallon jug and were asked to fill the 5-gallon jug with exactly 4 gallons. This problem generalizes that puzzle.
You have two jugs, A and B, and an infinite supply of water. There are three types of actions that you can use: (1) you can fill a jug, (2) you can empty a jug, and (3) you can pour from one jug to the other. Pouring from one jug to the other stops when the first jug is empty or the second jug is full, whichever comes first. For example, if A has 5 gallons and B has 6 gallons and a capacity of 8, then pouring from A to B leaves B full and 3 gallons in A.
A problem is given by a triple (Ca,Cb,N), where Ca and Cb are the capacities of the jugs A and B, respectively, and N is the goal. A solution is a sequence of steps that leaves exactly N gallons in jug B. The possible steps are
fill A
fill B
empty A
empty B
pour A B
pour B A
success
where "pour A B" means "pour the contents of jug A into jug B", and "success" means that the goal has been accomplished.
You may assume that the input you are given does have a solution.
输入
Input to your program consists of a series of input lines each defining one puzzle. Input for each puzzle is a single line of three positive integers: Ca, Cb, and N. Ca and Cb are the capacities of jugs A and B, and N is the goal. You can assume 0 < Ca <= Cb and N <= Cb <=1000 and that A and B are relatively prime to one another.
输出
Output from your program will consist of a series of instructions from the list of the potential output lines which will result in either of the jugs containing exactly N gallons of water. The last line of output for each puzzle should be the line "success". Output lines start in column 1 and there should be no empty lines nor any trailing spaces.
样例输入
3 7 1
9 32 6
样例输出
fill B
pour B A
empty A
pour B A
success
fill B
pour B A
empty A
pour B A
empty A
pour B A
empty A
pour B A
fill B
pour B A
empty A
pour B A
empty A
pour B A
empty A
pour B A
empty A
pour B A
fill B
pour B A
empty A
pour B A
empty A
pour B A
success分析：原题应该是这个：UVA - 571 ，但是原题可以过的代码在Codeup上就是过不了。可能是因为codeup上只有检查到容器B的水为n，才符合要求。
首先应该可以看出来，两个壶的条件下，倒水的过程实际上就是求最大公约数的更相减损术。只要目标值是两个壶容量的最大公约数的倍数，就一定有解。如果不考虑步骤，只要一个壶一直往另一个壶倒水，总能够达到要求。
#include<algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <cstdio>
#include <queue>
#include <stack>
#include <ctime>
#include <cmath>
#include <map>
#include <set>
#define INF 0x3f3f3f3f
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,a) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#a<<"="<<(a)<<endl
#define db5(x,y,z,a,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#a<<"="<<(a)<<", "<<#r<<"="<<(r)<<endl
using namespace std;
typedef struct node
{
int cnt_a,cnt_b,fat,op;
}node;
void print(node a)
{
// db4(a.cnt_a,a.cnt_b,a.fat,a.op);
if(a.op==1)printf("fill A\n");
else if(a.op==2)printf("fill B\n");
else if(a.op==3)printf("empty A\n");
else if(a.op==4)printf("empty B\n");
else if(a.op==5)printf("pour A B\n");
else if(a.op==6)printf("pour B A\n");
return;
}
int main(void)
{
#ifdef test
freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
clock_t start=clock();
#endif //test
int ca,cb,n;
while(~scanf("%d%d%d",&ca,&cb,&n))
{
if(n==0)
{
printf("success\n");continue;
}
int flag[ca+5][cb+5],l=0,r=1;
for(int i=0;i<=ca;++i)
for(int j=0;j<=cb;++j)
flag[i][j]=0;
vector<node>sta;
node temp;
temp.cnt_a=temp.cnt_b=0,temp.fat=-1,temp.op=0;
sta.push_back(temp);
flag[0][0]=1;
while(l<r)
{
int ll=l,rr=r,f=0,index=-1;
for(int i=ll;i<rr;++i)
{
// db3(sta[i].cnt_a,sta[i].cnt_b,i);
// if(flag[sta[i].cnt_a][sta[i].cnt_b]==1)continue;
// else flag[sta[i].cnt_a][sta[i].cnt_b]=1;
if(sta[i].cnt_b==n)
{
f=1;index=i;break;
}
if(sta[i].cnt_a<ca)//fill A
{
temp.cnt_a=ca;temp.cnt_b=sta[i].cnt_b;temp.fat=i,temp.op=1;
if(!flag[temp.cnt_a][temp.cnt_b])flag[temp.cnt_a][temp.cnt_b]=1,sta.push_back(temp),r++;
}
if(sta[i].cnt_b<cb)//fill B
{
temp.cnt_b=cb;temp.cnt_a=sta[i].cnt_a;temp.fat=i,temp.op=2;
if(!flag[temp.cnt_a][temp.cnt_b])flag[temp.cnt_a][temp.cnt_b]=1,sta.push_back(temp),r++;
}
if(sta[i].cnt_a>0)//empty A
{
temp.cnt_a=0;temp.cnt_b=sta[i].cnt_b;temp.fat=i,temp.op=3;
if(!flag[temp.cnt_a][temp.cnt_b])flag[temp.cnt_a][temp.cnt_b]=1,sta.push_back(temp),r++;
}
if(sta[i].cnt_b>0)//empty B
{
temp.cnt_b=0;temp.cnt_a=sta[i].cnt_a;temp.fat=i,temp.op=4;
if(!flag[temp.cnt_a][temp.cnt_b])flag[temp.cnt_a][temp.cnt_b]=1,sta.push_back(temp),r++;
}
if(sta[i].cnt_b<cb&&sta[i].cnt_a>0)//pour A B
{
if(cb-sta[i].cnt_b>=sta[i].cnt_a)temp.cnt_b=sta[i].cnt_b+sta[i].cnt_a,temp.cnt_a=0;
else temp.cnt_a=sta[i].cnt_a-cb+sta[i].cnt_b,temp.cnt_b=cb;
temp.fat=i,temp.op=5;
if(!flag[temp.cnt_a][temp.cnt_b])flag[temp.cnt_a][temp.cnt_b]=1,sta.push_back(temp),r++;
}
if(sta[i].cnt_b>0&&sta[i].cnt_a<ca)//pour B A
{
if(ca-sta[i].cnt_a>=sta[i].cnt_b)temp.cnt_a=sta[i].cnt_a+sta[i].cnt_b,temp.cnt_b=0;
else temp.cnt_b=sta[i].cnt_b-ca+sta[i].cnt_a,temp.cnt_a=ca;
temp.fat=i,temp.op=6;
if(!flag[temp.cnt_a][temp.cnt_b])flag[temp.cnt_a][temp.cnt_b]=1,sta.push_back(temp),r++;
}
}
l=rr;
// db4(l,r,f,index);
if(f==1)
{
stack<node>ans;
while(index!=-1)
{
// db1(index);
ans.push(sta[index]);index=sta[index].fat;
}
while(!ans.empty())
{
node cnt=ans.top();ans.pop();
print(cnt);
}
printf("success\n");
break;
}
}
// db3(ca,cb,n);
}
#ifdef test
clockid_t end=clock();
double endtime=(double)(end-start)/CLOCKS_PER_SEC;
printf("\n\n\n\n\n");
cout<<"Total time:"<<endtime<<"s"<<endl; //s为单位
cout<<"Total time:"<<endtime*1000<<"ms"<<endl; //ms为单位
#endif //test
return 0;
}