POJ

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题目链接：http://poj.org/problem?id=3414点击打开链接

Pots

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 17983 Accepted: 7603 Special Judge

Description

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.

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

6FILL(2)POUR(2,1)DROP(1)POUR(2,1)FILL(2)POUR(2,1)

Source

Northeastern Europe 2002, Western Subregion

跟hdu的非常可乐差不多主要是记录比较麻烦

每一步都记录之前的步数然后递推往后

也可以只记录每个步数的前驱然后最后放到一个栈倒序输出

#include <iostream>#include <queue>#include <stdio.h>#include <stdlib.h>#include <stack>#include <limits.h>#include <string>#include <string.h>#include <vector>#include <set>#include <map>#include <algorithm>#include <math.h>using namespace std;int n,m,k;struct xjy{    int nv;    int mv;    int type;    char num;    int step;    int pre;};struct book{    int x;    int y;};pair <int ,int >check;set<pair<int ,int > > s;vector <xjy> q;vector <xjy> ::iterator it;int iit;vector<xjy> aans;xjy ans;void bfs(){    xjy beg;    beg.nv=0;    beg.mv=0;    beg.step=0;    beg.num='k';    beg.pre=-1;    check.first=beg.nv;    check.second=beg.mv;    q.push_back(beg);    s.insert(check);    it=q.begin();    while(iit!=q.size())    {        xjy mid;        mid=q[iit];        iit++;        if(mid.mv==k||mid.nv==k)            {                ans=mid;                break;            }        xjy midmid;        midmid=mid;        //1.n        if(midmid.nv!=n)        {            midmid.nv=n;            midmid.type=1;            midmid.num='n';            midmid.step++;            midmid.pre=iit-1;            check.first=midmid.nv;            check.second=midmid.mv;            if(!s.count(check))            {                q.push_back(midmid);                s.insert(check);            }        }        //1.m        midmid=mid;        if(midmid.mv!=m)        {            midmid.mv=m;            midmid.type=1;            midmid.num='m';            midmid.step++;            midmid.pre=iit-1;            check.first=midmid.nv;            check.second=midmid.mv;            if(!s.count(check))            {                q.push_back(midmid);                s.insert(check);            }        }        //2.m        midmid=mid;        if(midmid.nv!=n&&midmid.mv!=0)        {            if(midmid.mv>(n-midmid.nv))            {                midmid.mv=midmid.mv-(n-midmid.nv);                midmid.nv=n;                midmid.type=2;                midmid.num='m';                midmid.step++;                midmid.pre=iit-1;                check.first=midmid.nv;                check.second=midmid.mv;                if(!s.count(check))                {                    q.push_back(midmid);                    s.insert(check);                }            }            else            {                midmid.nv+=midmid.mv;                midmid.mv=0;                midmid.type=2;                midmid.num='m';                midmid.step++;                midmid.pre=iit-1;                check.first=midmid.nv;                check.second=midmid.mv;                if(!s.count(check))                {                    q.push_back(midmid);                    s.insert(check);                }            }        }        //2.n        midmid=mid;        if(midmid.mv!=m&&midmid.nv!=0)        {            if(midmid.nv>(m-midmid.mv))            {                midmid.nv-=(m-midmid.mv);                midmid.mv=m;                midmid.type=2;                midmid.num='n';                midmid.step++;                midmid.pre=iit-1;                check.first=midmid.nv;                check.second=midmid.mv;                if(!s.count(check))                {                    q.push_back(midmid);                    s.insert(check);                }            }            else            {                midmid.mv+=midmid.nv;                midmid.nv=0;                midmid.type=2;                midmid.num='n';                midmid.step++;                midmid.pre=iit-1;                check.first=midmid.nv;                check.second=midmid.mv;                if(!s.count(check))                {                    q.push_back(midmid);                    s.insert(check);                }            }        }        //3.m        midmid=mid;        if(midmid.mv!=0)        {            midmid.mv=0;            midmid.type=3;            midmid.num='m';            midmid.step++;            midmid.pre=iit-1;            check.first=midmid.nv;            check.second=midmid.mv;            if(!s.count(check))            {                q.push_back(midmid);                s.insert(check);            }        }        //3.n        midmid=mid;        if(midmid.nv!=0)        {            midmid.nv=0;            midmid.type=3;            midmid.num='n';            midmid.step++;            midmid.pre=iit-1;            check.first=midmid.nv;            check.second=midmid.mv;            if(!s.count(check))            {                q.push_back(midmid);                s.insert(check);            }        }    }}int main(){    scanf("%d%d%d",&n,&m,&k);    //if((k&1)&&!(n&1)&&!(m&1))        //printf("impossible\n");    //else    {        bfs();        if(ans.step)        {            printf("%d\n",ans.step);            int ppre=ans.pre;            aans.push_back(ans);            while(ppre!=0)            {                aans.push_back(q[ppre]);                ppre=q[ppre].pre;            }            for(it=aans.end()-1;it>=aans.begin();it--)            {                if((*it).type==1)                {                    printf("FILL");                    printf("(");                    if((*it).num=='m')                        printf("2");                    else if((*it).num=='n')                        printf("1");                    printf(")\n");                }                else if((*it).type==2)                {                    printf("POUR");                    printf("(");                    if((*it).num=='m')                        printf("2,1");                    else if((*it).num=='n')                        printf("1,2");                    printf(")\n");                }                else if((*it).type==3)                {                    printf("DROP");                    printf("(");                    if((*it).num=='m')                        printf("2");                    else if((*it).num=='n')                        printf("1");                    printf(")\n");                }            }        }        else        {            printf("impossible\n");        }    }}

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