HDOJ 1495非常可乐

非常可乐原题链接

题目描述：
思路：
当可乐总量为奇数时，不可能均分，直接输出结果；将i被子倒入j杯中时，分两种情况，一种是能倒完，一种还有剩余

AC代码：

#include <iostream>
#include <queue>
#include <string.h>
using namespace std;

int N,M,S;
struct node
{
    int num[3];
    int steps;//次数
}now,nex;

int vis[101][101][101];
int rq[3];//最大容量

bool check(node x)//检查
{
    if((x.num[0] == x.num[1] && x.num[2] == 0 )||(x.num[0] == x.num[2] && x.num[1] == 0)||(x.num[1] == x.num[2] && x.num[0] == 0))
    {
        return true;
    }
    else return false;
}

void bfs()
{
    queue<node> Q;
    memset(vis,0,sizeof(vis));
    now.num[0] = S;
    now.num[1] = 0;
    now.num[2] = 0;
    vis[S][0][0] = 1;//记录当前的情况已经访问了
    now.steps = 0;//记录次数
    Q.push(now);
    while(!Q.empty())
    {
        now = Q.front();
        Q.pop();
        if(check(now))
        {
            printf("%d\n",now.steps);
            return;
        }
        for(int i = 0; i < 3; i++)
        {
            for(int j = 0; j < 3; j++)
            {
                if(i == j)  continue;
                if(now.num[i] <= rq[j] - now.num[j])//能倒完
                {
                    nex.num[j] = now.num[j] + now.num[i];
                    nex.num[i] = 0;
                }
                else//不能倒完
                {
                    nex.num[i] = now.num[i] - (rq[j] - now.num[j]);
                    nex.num[j] = rq[j];
                }
                //另外一个不变
                for(int k = 0; k < 3; k++)
                {
                    if(k == i || k == j)
                    {
                        continue;
                    }
                    nex.num[k] = now.num[k];
                }
                //步数加一，注意别写成now.steps++!!!
                nex.steps = now.steps + 1;
                if(vis[nex.num[0]][nex.num[1]][nex.num[2]]==0){
                    vis[nex.num[0]][nex.num[1]][nex.num[2]]=1;
                    Q.push(nex);
                }
            }
        }
    }
    printf("NO\n");
    return;
}
int main()
{
    while(scanf("%d%d%d",&S,&N,&M)!=EOF && S!=0 && N!=0 && M!=0)
    {
        
        if(S / 2 == 1)
        {
            printf("NO\n");
        }
        else{
            rq[0] = S;
            rq[1] = N;
            rq[2] = M;
            bfs();
        }
    }
    return 0;
}