HDU1495 非常可乐（BFS）

　　刚拿到这题时很纠结，没有思路。后来想了一天，问了问高手终于把思路整理出来。

思路：三个瓶子，无非就是6种倒法，按杯子体积从大到小定义为s, m, n。则6种倒法分别是：s -> m, s -> n, m -> n, m -> s,  n -> m, n -> s；

加一个标记变量，标记当前状态已经出现过。将没出现过的状态入队列，剩下的就是bfs的事了。。。

My Code:

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
const int N = 100000;
struct coco
{
    int s;
    int m;
    int n;
    int step;
}q[N];

int s, m, n;
int vis[101][101];

int bfs()
{
    coco p, t;
    int cnt, r = 0, f = 0;
    q[0].s = s;
    q[0].m = 0;
    q[0].n = 0;
    q[0].step = 0;
    vis[m][n] = 1;
    r++;
    while(f < r)
    {
        p = q[f++];
        cnt = 0;
        if(p.s == s/2)    cnt++;
        if(p.m == s/2)    cnt++;
        if(p.n == s/2)    cnt++;
        if(cnt == 2)        //satisfied the term
            return p.step;    
        if(p.s)
        {
            t = p;
            if(t.m < m)    //s -> m
            {
                if(t.m + t.s > m)
                {
                    t.s -= (m-t.m);
                    t.m = m;
                }
                else
                {
                    t.m += t.s;
                    t.s = 0;
                }
                t.step++;
                if(!vis[p.m][p.n])
                {
                    vis[p.m][p.n] = 1;
                    q[r++] = t;
                }
            }
            t = p;
            if(t.n < n)    //s -> n
            {
                if(t.n + t.s > n)
                {
                    t.s -= (n-t.n);
                    t.n = n;
                }
                else
                {
                    t.m += t.s;
                    t.s = 0;
                }
                t.step++;
                if(!vis[t.m][t.n])
                {
                    vis[t.m][t.n] = 1;
                    q[r++] = t;
                }
            }
        }
        if(p.m)
        {
            t = p;    //m -> n
            if(t.n < n)
            {
                if(t.n + t.m > n)
                {
                    t.m -= (n-t.n);
                    t.n = n;
                }
                else
                {
                    t.n += t.m;
                    t.m = 0;
                }
                t.step ++;
                if(!vis[t.m][t.n])
                {
                    vis[t.m][t.n] = 1;
                    q[r++] = t;
                }
            }

            t = p;    // m -> s
            t.s += t.m;
            t.m = 0;
            t.step ++;
            if(!vis[t.m][t.n])
            {
                vis[t.m][t.n] = 1;
                q[r++] = t;
            }
        }
        if(p.n)
        {
            t = p;        // n -> m
            if(t.n + t.m > m)
            {
                t.n -= (m - t.m);
                t.m = m;
            }
            else
            {
                t.m += t.n;
                t.n = 0;
            }
            t.step ++;
            if(!vis[t.m][t.n])
            {
                vis[t.m][t.n] = 1;
                q[r++] = t;
            }

            t = p;        //n -> s
            t.s += t.n;
            t.n = 0;
            t.step ++;
            if(!vis[t.m][t.n])
            {
                vis[t.m][t.n] = 1;
                q[r++] = t;
            }
        }
    }
    return 0;
}

int main()
{
    //freopen("data.in", "r", stdin);

    int ans;
    while(scanf("%d%d%d", &s, &n, &m), s||n||m)
    {
        memset(vis, 0, sizeof(vis));
        if(m > n)
        {
            m = m+n;
            n = m-n;
            m = m-n;
        }
        if(s&1)        //if 's' is odd
        {
            printf("NO\n");
            continue;
        }
        ans = bfs();
        if(ans)
            printf("%d\n", ans);
        else
            printf("NO\n");
    }
    return 0;
}