非常可乐

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非常可乐
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit

Status

Practice

HDU 1495
Description
大家一定觉的运动以后喝可乐是一件很惬意的事情，但是seeyou却不这么认为。因为每次当seeyou买了可乐以后，阿牛就要求和seeyou一起分享这一瓶可乐，而且一定要喝的和seeyou一样多。但seeyou的手中只有两个杯子，它们的容量分别是N 毫升和M 毫升可乐的体积为S （S<101）毫升　(正好装满一瓶) ，它们三个之间可以相互倒可乐 (都是
没有刻度的，且 S==N+M，101＞S＞0，N＞0，M＞0) 。聪明的ACMER你们说他们能平分吗？如果能请输出倒可乐的最少的次数，如果不能输出"NO"。

Input
三个整数 : S 可乐的体积 , N 和 M是两个杯子的容量，以"0 0 0"结束。

Output
如果能平分的话请输出最少要倒的次数，否则输出"NO"。

Sample Input
7 4 3
4 1 3
0 0 0

Sample Output
NO

#include <cstdio>#include <cstring>#include <queue>using namespace std;typedef struct rot{    int s, n, m;    int num;}rot;int s, n, m;bool pos[101][101][101];bool judge(rot x){    if( pos[x.s][x.n][x.m] || x.s<0||x.s>s || x.m<0||x.m>m ||x.n<0||x.n>n)    return false;    return true;}int bfs(){    rot now, next;    queue<rot>Q;    now.s=s; now.n=0; now.m=0; now.num=0;    Q.push( now );    pos[now.s][now.n][now.m] = true;    while( !Q.empty() ){        now=Q.front(); Q.pop();        if(now.s==s/2 && now.m==0)    return now.num;        //s->m;        if( now.s>(m-now.m) ){            next.s=now.s-(m-now.m);    next.m=m;   next.n=now.n;  next.num=now.num+1;    if(judge(next)){ Q.push(next);   pos[next.s][next.n][next.m]=true; }        }        else if(now.s!=0){            next.s=0;   next.m=now.m+now.s;   next.n=now.n;   next.num=now.num+1;     if(judge(next)){ Q.push(next);   pos[next.s][next.n][next.m]=true; }        }        //s->n;        if( now.s>(n-now.n) ){            next.s=now.s-(n-now.n);    next.n=n;   next.m=now.m;  next.num=now.num+1;    if(judge(next)){ Q.push(next);   pos[next.s][next.n][next.m]=true; }        }        else if(now.s!=0){            next.s=0;   next.n=now.n+now.s;   next.m=now.m;   next.num=now.num+1;     if(judge(next)){ Q.push(next);   pos[next.s][next.n][next.m]=true; }        }        //n->s;        if( now.n>(s-now.s) ){            next.n=now.n-(s-now.s);    next.s=s;   next.m=now.m;  next.num=now.num+1;    if(judge(next)){ Q.push(next);   pos[next.s][next.n][next.m]=true; }        }        else if(now.n!=0){            next.n=0;   next.s=now.s+now.n;   next.m=now.m;   next.num=now.num+1;     if(judge(next)){ Q.push(next);   pos[next.s][next.n][next.m]=true; }        }        //n->m;        if( now.n>(m-now.m) ){            next.n=now.n-(m-now.m);    next.m=m;   next.s=now.s;  next.num=now.num+1;    if(judge(next)){ Q.push(next);   pos[next.s][next.n][next.m]=true; }        }        else if(now.n!=0){            next.n=0;   next.m=now.m+now.n;   next.s=now.s;   next.num=now.num+1;     if(judge(next)){ Q.push(next);   pos[next.s][next.n][next.m]=true; }        }        //m->s;        if( now.m>(s-now.s) ){            next.m=now.m-(s-now.s);    next.s=s;   next.n=now.n;  next.num=now.num+1;    if(judge(next)){ Q.push(next);   pos[next.s][next.n][next.m]=true; }        }        else if(now.m!=0){            next.m=0;   next.s=now.s+now.m;   next.n=now.n;   next.num=now.num+1;     if(judge(next)){ Q.push(next);   pos[next.s][next.n][next.m]=true; }        }        //m->n;        if( now.m>(n-now.n) ){            next.m=now.m-(n-now.n);    next.n=n;   next.s=now.s;  next.num=now.num+1;    if(judge(next)){ Q.push(next);   pos[next.s][next.n][next.m]=true; }        }        else if(now.m!=0){            next.m=0;   next.n=now.n+now.m;   next.s=now.s;   next.num=now.num+1;     if(judge(next)){ Q.push(next);   pos[next.s][next.n][next.m]=true; }        }    }    return -1;}int main(){    int ans;    while( scanf("%d%d%d", &s,&n,&m)!=-1 ){        if( s==0 && n==0 && m==0 )  break;        if(m>n) { int flag=m; m=n; n=flag; }        memset(pos, false, sizeof(pos));        if(s%2==0)            ans = bfs();        else ans=-1;        if(ans==-1)   printf("NO\n");        else   printf("%d\n", ans);    }    return 0;}

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