标签：des style color io os ar for strong sp
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 10071 | Accepted: 4237 | 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)
思路：一共有6种操作：把A中水倒掉，把A加满，把B里的水倒入A中；B和A类似。
罐子最大容积为100，设一个常量N=100， 开一个二维数组记录状态变化的值。
1、从水龙头往A里加水t,记录-t,
2、从水龙头往B里加水t,记录-t-N,
3、从B里面往A加水t，记录t
4、从A里面往B加水t，记录N+t
5、把A里水倒掉，记录2*N+t,(A原有水t)
6、把B里水倒掉，记录3*N+t,(B原有水t)
#include<stdio.h>
#include<queue>
#include<map>
#include<string>
#include<string.h>
using namespace std;
#define N 105
const int inf=0x1f1f1f1f;
int a,b,c,flag;
int mark[N][N];
struct node
{
int x,y,t;
friend bool operator<(node a,node b)
{
return a.t>b.t;
}
};
void prif(int x,int y) //递归输出路径
{
if(x==0&&y==0)
return ;
if(mark[x][y]>3*N)
{
prif(x,mark[x][y]-3*N);
printf("DROP(2)\n");
}
else if(mark[x][y]>2*N)
{
prif(mark[x][y]-2*N,y);
printf("DROP(1)\n");
}
else if(mark[x][y]>N)
{
int tmp=mark[x][y]-N;
prif(x+tmp,y-tmp);
printf("POUR(1,2)\n");
}
else if(mark[x][y]>0)
{
int tmp=mark[x][y];
prif(x-tmp,y+tmp);
printf("POUR(2,1)\n");
}
else if(mark[x][y]>-N)
{
int tmp=-mark[x][y];
prif(x-tmp,y);
printf("FILL(1)\n");
}
else if(mark[x][y]<-N)
{
int tmp=N+mark[x][y];
prif(x,y+tmp);
printf("FILL(2)\n");
}
}
void bfs()
{
priority_queue<node>q;
node cur,next;
mark[0][0]=inf; //该状态只能出现一次，赋值为inf防止干扰其他值
mark[a][0]=-a;
mark[0][b]=-b-N;
cur.t=1;
cur.x=a;
cur.y=0;
q.push(cur);
cur.x=0;
cur.y=b;
q.push(cur);
while(!q.empty())
{
cur=q.top();
q.pop();
next.t=cur.t+1;
if(cur.x==c||cur.y==c)
{
flag=1;
printf("%d\n",cur.t);
prif(cur.x,cur.y);
return ;
}
if(cur.x<a) //向A加水
{
int tmp=a-cur.x;
next.y=cur.y;
next.x=a; //来自水龙头的水
if(!mark[next.x][next.y])
{
mark[next.x][next.y]=-tmp;
q.push(next);
}
if(cur.y>0) //来自B的水
{
int tmp=min(cur.y,a-cur.x);
next.x=cur.x+tmp;
next.y=cur.y-tmp;
if(!mark[next.x][next.y])
{
mark[next.x][next.y]=tmp;
q.push(next);
}
}
}
if(cur.y<b) //向B加水
{
int tmp=b-cur.y;
next.x=cur.x;
next.y=b; //来自水龙头的水
if(!mark[next.x][next.y])
{
mark[next.x][next.y]=-tmp-N;
q.push(next);
}
if(cur.x>0) //来自A的水
{
int tmp=min(cur.x,b-cur.y);
next.y=cur.y+tmp;
next.x=cur.x-tmp;
if(!mark[next.x][next.y])
{
mark[next.x][next.y]=tmp+N;
q.push(next);
}
}
}
if(cur.x>0) //倒掉水
{
int tmp=cur.x;
next.x=0;
next.y=cur.y;
if(!mark[next.x][next.y])
{
mark[next.x][next.y]=2*N+tmp;
q.push(next);
}
}
if(cur.y>0)
{
int tmp=cur.y;
next.y=0;
next.x=cur.x;
if(!mark[next.x][next.y])
{
mark[next.x][next.y]=3*N+tmp;
q.push(next);
}
}
}
}
int main()
{
while(scanf("%d%d%d",&a,&b,&c)!=-1)
{
memset(mark,0,sizeof(mark));
flag=0;
bfs();
if(!flag)
printf("impossible\n");
}
return 0;
}
标签：des style color io os ar for strong sp
原文地址：http://blog.csdn.net/u011721440/article/details/39803613
