HDU 1495 非常可乐

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

3

 

题意：中文题，不用多说，三个杯子可以互相倒可乐，然而并没有刻度，那么只能依靠杯子的总容量来算了，首先确定当一个杯子有可乐时就可以向其他两个杯子倒可乐了，那么只要判断另外两个杯子的状态就行了，每次像其他杯子倒水都算作一次（可以确定有六种可能的状态了），剩下的就是一个BFS了，就是有点麻烦，代码可长。。。。

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

const int maxn=110;

struct node
{
    int s, n, m, k; ///s，n，m分别表示可乐瓶和两个杯子中可乐的容量，k表示倒可乐的次数
};
int S, N, M, vis[maxn][maxn][maxn];

int BFS()
{
    node now, next;
    queue<node>Q;

    now.s = S;
    now.n = now.m = now.k = 0;

    Q.push(now);
    vis[now.s][now.n][now.m] = 1;

    while (!Q.empty())
    {
        now = Q.front(); Q.pop();

        if ((now.s==S/2 && now.n==S/2) || (now.s==S/2 && now.m==S/2) || (now.n==S/2 && now.m==S/2))
            return now.k; ///题中要求只要能平分就行，所以不管哪两个杯子装的是平分的可乐都行
        ///接下来就是三个杯子分别有可乐的情况，每种情况中又分别判断了另两个杯子的状态，所以一共是六种
        if (now.s != 0) 
        {
            if (now.s <= N-now.n)
            {
                next.s = 0;
                next.n = now.n+now.s;
                next.m = now.m;
                next.k = now.k+1;
            }
            else
            {
                next.s = now.s-(N-now.n);
                next.n = N;
                next.m = now.m;
                next.k = now.k+1;
            }
            if (!vis[next.s][next.n][next.m])
            {
                vis[next.s][next.n][next.m] = 1;
                Q.push(next);
            }

            if (now.s <= M-now.m)
            {
                next.s = 0;
                next.n = now.n;
                next.m = now.m+now.s;
                next.k = now.k+1;
            }
            else
            {
                next.s = now.s-(M-now.m);
                next.n = now.n;
                next.m = M;
                next.k = now.k+1;
            }
            if (!vis[next.s][next.n][next.m])
            {
                vis[next.s][next.n][next.m] = 1;
                Q.push(next);
            }
        }

        if (now.n != 0)
        {
            if (now.n <= S-now.s)
            {
                next.s = now.s+now.n;
                next.n = 0;
                next.m = now.m;
                next.k = now.k+1;
            }
            else
            {
                next.s = S;
                next.n = now.n-(S-now.s);
                next.m = now.m;
                next.k = now.k+1;
            }
            if (!vis[next.s][next.n][next.m])
            {
                vis[next.s][next.n][next.m] = 1;
                Q.push(next);
            }

            if (now.n <= M-now.m)
            {
                next.s = now.s;
                next.n = 0;
                next.m = now.m+now.n;
                next.k = now.k+1;
            }
            else
            {
                next.s = now.s;
                next.n = now.n-(M-now.m);
                next.m = M;
                next.k = now.k+1;
            }
            if (!vis[next.s][next.n][next.m])
            {
                vis[next.s][next.n][next.m] = 1;
                Q.push(next);
            }
        }

        if (now.m != 0)
        {
            if (now.m <= S-now.s)
            {
                next.s = now.s+now.m;
                next.n = now.n;
                next.m = 0;
                next.k = now.k+1;
            }
            else
            {
                next.s = S;
                next.n = now.n;
                next.m = now.m-(S-now.s);;
                next.k = now.k+1;
            }
            if (!vis[next.s][next.n][next.m])
            {
                vis[next.s][next.n][next.m] = 1;
                Q.push(next);
            }

            if (now.m <= N-now.n)
            {
                next.s = now.s;
                next.n = now.m+now.n;
                next.m = 0;
                next.k = now.k+1;
            }
            else
            {
                next.s = now.s;
                next.n = N;
                next.m = now.m-(N-now.n);
                next.k = now.k+1;
            }
            if (!vis[next.s][next.n][next.m])
            {
                vis[next.s][next.n][next.m] = 1;
                Q.push(next);
            }
        }
    }

    return -1;
}

int main ()
{
    int ans;

    while (scanf("%d%d%d", &S, &N, &M), S+N+M)
    {
        memset(vis, 0, sizeof(vis));

        if (S % 2 != 0) printf("NO
"); ///当然可乐的容量原本就是奇数，是肯定不会被平分的
        else
        {
            ans = BFS();

            if (ans == -1) printf("NO
");
            else printf("%d
", ans);
        }
    }

    return 0;
}