非常可乐 hdu 1495

此题需要注意的是。最后的判断条件应是。。S == M + N ..

刚开始把判断条件搞错了。认为只要任意两杯相加与第三杯相等即可。。

悲剧啊。改了我一个多时侯。改来改去还是测试数据还是2.。。

最后把测试数据手算下才知道。。

判断条件写错了。。

代码写得有点差。。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <queue>

using namespace std;
int S, N, M, t;

struct node
{
    int s, n, m, t;
    bool operator < (const node &A) const
    {
        return A.t < t;
    }
    
}NODE;

int c1[120];
int c2[120];
int c3[120];


int main( )
{
    while (scanf("%d%d%d", &S, &N, &M) , S || N || M)
    {   
        int r, s1, n1, m1, f1 = 0, n, m, s, t, t1;
        memset(c1, 0, sizeof(c1));
        memset(c2, 0, sizeof(c2));
        memset(c3, 0, sizeof(c3));
        
        priority_queue<node>q;
        
        NODE.s = S, NODE.t = 0, NODE.m = 0, NODE.n = 0;
        q.push(NODE);
        c1[S] = 1, c2[0] = 1, c3[0] = 1;
        while ( !q.empty( ))
        {
            NODE =  q.top( );
            q.pop( );
            s = NODE.s, t = NODE.t;
            m = NODE.m, n = NODE.n;
             
            if( ( s == m + n  ) )
                {  
                    f1  = 1;
                    printf("%d\n",t);
                    break;
                }
            //对大杯子的操作,倒在M里 
            if(s > 0)
             {
             if ( (s - (M - m) ) >= 0)
                 m1 = M, s1 = s - (M - m);
             else
                 m1 =  s + m, s1 = 0;
             n1 = n;
             NODE.s = s1, NODE.t = t + 1, NODE.m = m1, NODE.n = n1;
             if ( s1 >= 0 && m1 <= M && n1 <= N && (!c1[s1] || !c2[m1] || !c3[n1]) )
             {
                        c1[s1] = c2[m1] = c3[n1] = 1;
                        q.push(NODE);
                    //    printf("1:(%d ,%d, %d)->%d\n",s1, n1, m1, t + 1);
             }
             }
             //对大杯子的操作,倒在N里
             if(s > 0)
             {
             if ( (s - (N - n) ) >= 0 )
                  s1 = s- (N - n), n1 = N;
             else
                   s1 = 0, n1 = n + s;
             m1 = m;
             NODE.s = s1, NODE.t = t + 1, NODE.n = n1, NODE.m = m1;
             if ( s1 >= 0 && m <= M && n1 <= N && (!c1[s1] || !c2[m1] || !c3[n1] ) )
             {
                        c1[s1] = 1;
                        c2[m1]  = 1;
                        c3[n1] = 1;
                        q.push(NODE);
                    //    printf("2:(%d ,%d, %d)->%d\n",s1, n1, m1, t  + 1);
             }
             }
             //对N的操作，倒在S大杯子里 
             if(n > 0)
             {
             if (n <= (S- s) )
                     n1 = 0, s1 = s + n;
             else
                     n1 = n - (S -s), s1 = S;
             m1 = m;
             NODE.s = s1, NODE.t = t + 1, NODE.n = n1, NODE.m = m1;
             if ( s1 >= 0 && m <= M && n1 <= N && (!c1[s1] || !c2[m1] || !c3[n1] ))
             {
                    
                        c1[s1] = 1;
                        c2[m1] = 1;
                        c3[n1] = 1;
                        q.push(NODE);
                    //    printf("3:(%d ,%d, %d)->%d\n",s1, n1, m1, t + 1);
             }
             }
             //对N的操作，倒在M杯子里 
             if( n > 0)
             {
             if (n <= (M- m) )
                     n1 = 0, m1 = m + n;
             else
                     n1 = n - (M -m), m1 = M;
             NODE.m = m1, NODE.t = t + 1, NODE.n = n1;
             NODE.s = s;
             if ( s >= 0 && m1 <= M && n1 <= N && (!c1[s] || !c2[m1] || !c3[n1]))             {
                        c1[s] = 1;
                        c2[m1] = 1;
                        c3[n1] = 1;
                        q.push(NODE);
                        //printf("4:(%d ,%d, %d)->%d\n",s, n1, m1, t + 1);
             }
             }
              //对M的操作，倒在S杯子里 
             if( m > 0)
             {
             if (m <= (S- s) )
                     m1 = 0, s1 = s  + m;
             else
                     m1 = m - (S -s), s1 = S;
             NODE.m = m1, NODE.t = t + 1, NODE.s = s1, NODE.n = n;
             if ( s1 >= 0 && m1 <= M && n <= N  && !c1[s1] || !c2[m1] || !c3[n])
             {
                        c1[s1] = 1;
                        c2[m1] = 1;
                        c3[n] = 1;
                        q.push(NODE);
                    //    printf("5:(%d ,%d, %d)->%d\n",s1, n1, m1, t);
             }
             }
              //对M的操作，倒在N杯子里杯子里 
             if( m > 0 )
             {
             if (m <= (N-n) )
                     m1 = 0, n1 = n + m;
             else
                     m1 = m - (N -n), n1 = N;
             NODE.m = m1, NODE.t = t + 1, NODE.n = n1;
             NODE.s=s;
             if ( s >= 0 && m1 <= M && n1 <= N && !c1[s] || !c2[m1] || !c3[n1])
             {
                        c1[s] = 1;
                        c2[m1] = 1;
                        c3[n1] = 1;
                        q.push(NODE);
                    //    printf("6:(%d ,%d, %d)->%d\n",s, n1, m1, t + 1);
             }
             }
             
        }
        if (!f1)
            puts("NO");
    }
     return 0;
    }