HDU 1495 非常可乐

#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\n"); ///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\n");
            else printf("%d\n", ans);
        }
    }

    return 0;
}