hdu 1495 非常可乐（广搜）

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题目地址
题目大意：给出三个数s，n，m，（s=m+n）初始态为s装满可乐，n和m为空，求是否能将可能均分，若能输出倒可乐的最小次数
解题思路：对于3个杯中每个杯子有可乐时用队列将状态加入队尾，且标记当前状态，直到队列中有均分的情况停止
#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <queue>#include <string>#include <map>#include <stack>#include <list>#include <set>using namespace std;const int maxn = 100+10;int visit[maxn][maxn][maxn];int s,n,m;struct Node{    int s,n,m,step;};int check(int x,int y,int z){    if(x==0 && y==z)    return 1;    if(y==0 && x==z)    return 1;    if(z==0 && x==y)    return 1;    return 0;}int bfs(){    queue<Node> q;    Node a,next;    a.s = s,a.n = 0,a.m = 0,a.step = 0;    visit[s][0][0] = 1;    q.push(a);    while(!q.empty())    {        a = q.front();        q.pop();        if(check(a.s,a.n,a.m))  return a.step;        if(a.n)//n还有        {            if(a.n>=s-a.s)//n能将s倒满，将n倒入s装满s            {                next = a;                next.n = next.n-(s-a.s);//剩余的n                next.s = s;//s满了                if(!visit[next.s][next.n][next.m])                {                    next.step = a.step+1;                    q.push(next);                    visit[next.s][next.n][next.m] = 1;                }            }            else //n不能将s倒满，将n全部倒入s中            {                next = a;                next.s = next.n+next.s;                next.n = 0;                if(!visit[next.s][next.n][next.m])                {                    next.step = a.step+1;                    q.push(next);                    visit[next.s][next.n][next.m] = 1;                }            }            if(a.n>=m-a.m)//n能将m倒满，将n倒入m装满m            {                next = a;                next.n = next.n-(m-a.m);                next.m =  m;                if(!visit[next.s][next.n][next.m])                {                    next.step = a.step+1;                    q.push(next);                    visit[next.s][next.n][next.m] = 1;                }            }            else//n不能将m倒满，将n全部倒入m中            {                next = a;                next.m = next.n+next.m;                next.n = 0;                if(!visit[next.s][next.n][next.m])                {                    next.step = a.step+1;                    q.push(next);                    visit[next.s][next.n][next.m] = 1;                }            }        }        if(a.m)//m还有        {            if(a.m>=s-a.s)//m能将s倒满，将m倒入s装满s            {                next = a;                next.m = next.m-(s-a.s);                next.s = s;                if(!visit[next.s][next.n][next.m])                {                    next.step = a.step+1;                    q.push(next);                    visit[next.s][next.n][next.m] = 1;                }            }            else //m不能将s倒满，将m全部倒入s中            {                next = a;                next.s = next.m+next.s;                next.m = 0;                if(!visit[next.s][next.n][next.m])                {                    next.step = a.step+1;                    q.push(next);                    visit[next.s][next.n][next.m] = 1;                }            }            if(a.m>=n-a.n)//m能将n倒满，将m倒入n装满n            {                next = a;                next.m = next.m-(n-a.n);                next.n =  n;                if(!visit[next.s][next.n][next.m])                {                    next.step = a.step+1;                    q.push(next);                    visit[next.s][next.n][next.m] = 1;                }            }            else//m不能将n倒满，将m全部倒入n中            {                next = a;                next.n = next.n+next.m;                next.m = 0;                if(!visit[next.s][next.n][next.m])                {                    next.step = a.step+1;                    q.push(next);                    visit[next.s][next.n][next.m] = 1;                }            }        }        if(a.s)//s还有        {            if(a.s>=m-a.m)//s能将m倒满，将s倒入m装满m            {                next = a;                next.s = next.s-(m-a.m);                next.m = m;                if(!visit[next.s][next.n][next.m])                {                    next.step = a.step+1;                    q.push(next);                    visit[next.s][next.n][next.m] = 1;                }            }            else //S不能将m倒满，将s全部倒入m中            {                next = a;                next.m = next.m+next.s;                next.s = 0;                if(!visit[next.s][next.n][next.m])                {                    next.step = a.step+1;                    q.push(next);                    visit[next.s][next.n][next.m] = 1;                }            }            if(a.s>=n-a.n)//s能将n倒满，将s倒入n装满n            {                next = a;                next.s = next.s-(n-a.n);                next.n =  n;                if(!visit[next.s][next.n][next.m])                {                    next.step = a.step+1;                    q.push(next);                    visit[next.s][next.n][next.m] = 1;                }            }            else//s不能将n倒满，将s全部倒入n中            {                next = a;                next.n = next.n+next.s;                next.s = 0;                if(!visit[next.s][next.n][next.m])                {                    next.step = a.step+1;                    q.push(next);                    visit[next.s][next.n][next.m] = 1;                }            }        }    }    return 0;}int main(){    int ans;    while(scanf("%d%d%d",&s,&n,&m) != EOF)    {        if(!(s+n+m))    break;        if(s%2)        {            printf("NO\n");            continue;        }        memset(visit,0,sizeof(visit));        ans = bfs();        if(ans)            printf("%d\n",ans);        else            printf("NO\n");    }    return 0;}
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