标签：poj 3414 pots bfs 递归打印
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 10255 | Accepted: 4333 | 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)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#define N 100010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos ( -1.0 );
const double E = 2.718281828;
typedef long long ll;
const int INF = 1000010;
using namespace std;
int n,m,x;
string s[3]= {"FILL","DROP","POUR"};
bool vis[110][110];
struct node
{
int num,nex;///num==：1（FILL），2（DROP），3（POUR）
int t;///倒满,guang的瓶子
int x1,x2;///num=3时，由x1->x2
int A,B;///a,b状态
int step;///步数
} mp[1010];
queue<node>que;
int fro,nex;
bool bfs()
{
while(que.size())
que.pop();
memset(vis,0,sizeof vis);
node a;
fro=0,nex=1;
mp[fro].A=0,mp[fro].B=0;
mp[fro].nex=-1;
mp[fro].step=0;
que.push(mp[fro]);
vis[0][0]=1;
while(que.size())
{
a=que.front();
que.pop();
if(a.A==x||a.B==x)
{
printf("%d\n",a.step);
return 1;
}
mp[nex].nex=-1;
if(!vis[n][a.B])///倒满A
{
mp[nex].nex=fro;
mp[nex].A=n,mp[nex].B=a.B;
mp[nex].num=1;
mp[nex].t=1;
mp[nex].step=a.step+1;
que.push(mp[nex]);
vis[n][a.B]=1;
nex++;
}
if(!vis[a.A][m])///倒满B
{
mp[nex].nex=fro;
mp[nex].A=a.A,mp[nex].B=m;
mp[nex].num=1;
mp[nex].t=2;
mp[nex].step=a.step+1;
que.push(mp[nex]);
vis[a.A][m]=1;
nex++;
}
if(!vis[0][a.B])///倒光A
{
mp[nex].nex=fro;
mp[nex].A=0,mp[nex].B=a.B;
mp[nex].num=2;
mp[nex].t=1;
mp[nex].step=a.step+1;
que.push(mp[nex]);
vis[0][a.B]=1;
nex++;
}
if(!vis[a.A][0])///倒光B
{
mp[nex].nex=fro;
mp[nex].A=a.A,mp[nex].B=0;
mp[nex].num=2;
mp[nex].t=2;
mp[nex].step=a.step+1;
que.push(mp[nex]);
vis[a.A][m]=1;
nex++;
}
if(a.A>0)///A->B
{
int t=m-a.B;
if(a.A>t)//
{
mp[nex].A=a.A-t;
mp[nex].B=m;
}
else //
{
mp[nex].A=0;
mp[nex].B=a.A+a.B;
}
if(!vis[mp[nex].A][mp[nex].B])
{
mp[nex].num=3;
mp[nex].x1=1,mp[nex].x2=2;
mp[nex].nex=fro;
mp[nex].step=a.step+1;
que.push(mp[nex]);
vis[mp[nex].A][mp[nex].B]=1;
nex++;
}
}
if(a.B>0)///B->A
{
int t=n-a.A;
if(a.B>t)//
{
mp[nex].A=n;
mp[nex].B=a.B-t;
}
else //
{
mp[nex].A=a.A+a.B;
mp[nex].B=0;
}
if(!vis[mp[nex].A][mp[nex].B])
{
mp[nex].num=3;
mp[nex].x1=2,mp[nex].x2=1;
mp[nex].nex=fro;
mp[nex].step=a.step+1;
que.push(mp[nex]);
vis[mp[nex].A][mp[nex].B]=1;
nex++;
}
}
fro++;
}
return 0;
}
void print(int x)
{
if(mp[x].nex!=-1)
{
print(mp[x].nex);
if(mp[x].num==1)
printf("FILL(%d)\n",mp[x].t);
else if(mp[x].num==2)
printf("DROP(%d)\n",mp[x].t);
else
printf("POUR(%d,%d)\n",mp[x].x1,mp[x].x2);
}
}
int main()
{
while(cin>>n>>m>>x)
{
if(!bfs())
cout<<"impossible"<<endl;
else
{
print(fro);
}
}
return 0;
}
标签：poj 3414 pots bfs 递归打印
原文地址：http://blog.csdn.net/acm_baihuzi/article/details/42440877
