HDU - 1495 非常可乐（bfs）

Posted 2020-10-30 多一份不为什么的坚持

非常可乐

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20347    Accepted Submission(s): 8248

Problem 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

 

Author

seeyou

 

Source

“2006校园文化活动月”之“校庆杯”大学生程序设计竞赛暨杭州电子科技大学第四届大学生程序设计竞赛

Recommend

LL

Statistic | Submit | Discuss | Note

思路：a可以往b中倒水，可以往c倒。。。。总共有六种情况，BFS广度搜索就行了

注意搜索的时候要标记一下状态，用visit，不然会queue会超限。

就是代码有点多，要仔细。。。

 网上大佬用数学方法做的orz。。。。https://blog.csdn.net/v5zsq/article/details/52097459，看不懂啊。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <queue>
  4 #include <cstring>
  5 
  6 using namespace std;
  7 
  8 typedef struct info{
  9     int a,b,c;
 10     int countt;
 11 }cup;
 12 
 13 int s,m,n;
 14 cup c1;
 15 cup now,nextt;
 16 int visit[105][105][105];
 17 
 18 
 19 int bfs(cup c){
 20 //    while(!q.empty()){
 21 //        q.pop();
 22 //    }
 23     queue<cup> q;
 24     q.push(c);
 25     while(!q.empty()){
 26         now=q.front(); q.pop();
 27 
 28         if((now.a==s/2&&now.b==s/2)||(now.a==s/2&&now.c==s/2)||(now.b==s/2&&now.c==s/2)){
 29             return now.countt;
 30         }
 31         if(now.a!=0){//第一个杯子向其他两个倒水
 32             int vol=n-now.b;
 33             if(vol!=0){
 34                 if(now.a>=vol){
 35                     nextt.b=n;
 36                     nextt.a=now.a-vol;
 37                     nextt.c=now.c;
 38                     nextt.countt=now.countt+1;
 39                 }else{
 40                     nextt.b=now.b+now.a;
 41                     nextt.a=0;
 42                     nextt.c=now.c;
 43                     nextt.countt=now.countt+1;
 44                 }
 45                 if(!visit[nextt.a][nextt.b][nextt.c]){
 46                     visit[nextt.a][nextt.b][nextt.c]=1;
 47                     q.push(nextt);
 48                 }
 49             }
 50 
 51             vol=m-now.c;
 52             if(vol!=0){
 53                 if(now.a>=vol){
 54                     nextt.c=m;
 55                     nextt.a=now.a-vol;
 56                     nextt.b=now.b;
 57                     nextt.countt=now.countt+1;
 58                     q.push(nextt);
 59                 }else{
 60                     nextt.c=now.c+now.a;
 61                     nextt.a=0;
 62                     nextt.b=now.b;
 63                     nextt.countt=now.countt+1;
 64                 }
 65                 if(!visit[nextt.a][nextt.b][nextt.c]){
 66                     visit[nextt.a][nextt.b][nextt.c]=1;
 67                     q.push(nextt);
 68                 }
 69             }
 70         }
 71 
 72         if(now.b!=0){//第二个杯子忘其他两个倒水
 73             int vol=s-now.a;
 74             if(vol!=0){
 75                 if(now.b>=vol){
 76                     nextt.a=s;
 77                     nextt.b=now.b-vol;
 78                     nextt.c=now.c;
 79                     nextt.countt=now.countt+1;
 80                 }else{
 81                     nextt.a=now.a+now.b;
 82                     nextt.b=0;
 83                     nextt.c=now.c;
 84                     nextt.countt=now.countt+1;
 85                 }
 86                 if(!visit[nextt.a][nextt.b][nextt.c]){
 87                     visit[nextt.a][nextt.b][nextt.c]=1;
 88                     q.push(nextt);
 89                 }
 90             }
 91 
 92             vol=m-now.c;
 93             if(vol!=0){
 94                 if(now.b>=vol){
 95                     nextt.c=m;
 96                     nextt.a=now.a;
 97                     nextt.b=now.b-vol;
 98                     nextt.countt=now.countt+1;
 99                 }else{
100                     nextt.c=now.c+now.b;
101                     nextt.b=0;
102                     nextt.a=now.a;
103                     nextt.countt=now.countt+1;
104                 }
105                 if(!visit[nextt.a][nextt.b][nextt.c]){
106                     visit[nextt.a][nextt.b][nextt.c]=1;
107                     q.push(nextt);
108                 }
109             }
110         }
111 
112         if(now.c!=0){//第三个杯子忘其他两个倒水
113             int vol=s-now.a;
114             if(vol!=0){
115                 if(now.c>=vol){
116                     nextt.a=s;
117                     nextt.c=now.c-vol;
118                     nextt.b=now.b;
119                     nextt.countt=now.countt+1;
120                 }else{
121                     nextt.a=now.a+now.c;
122                     nextt.c=0;
123                     nextt.b=now.b;
124                     nextt.countt=now.countt+1;
125                 }
126                 if(!visit[nextt.a][nextt.b][nextt.c]){
127                     visit[nextt.a][nextt.b][nextt.c]=1;
128                     q.push(nextt);
129                 }
130             }
131 
132             vol=n-now.b;
133             if(vol!=0){
134                 if(now.c>=vol){
135                     nextt.b=n;
136                     nextt.c=now.c-vol;
137                     nextt.a=now.a;
138                     nextt.countt=now.countt+1;
139                 }else{
140                     nextt.b=now.b+now.c;
141                     nextt.c=0;
142                     nextt.a=now.a;
143                     nextt.countt=now.countt+1;
144                 }
145                 if(!visit[nextt.a][nextt.b][nextt.c]){
146                     visit[nextt.a][nextt.b][nextt.c]=1;
147                     q.push(nextt);
148                 }
149             }
150         }
151     }
152     return 0;
153 }
154 
155 int main()
156 {
157     while(~scanf("%d %d %d",&s,&n,&m)){
158         memset(visit,0,sizeof(visit));
159         if(s==0&&m==0&&n==0){break;}
160         if(m==n){
161             printf("1\n");
162         }else if(s&1==1){
163             printf("NO\n");
164         }else{
165             c1.a=s; c1.b=0; c1.c=0; c1.countt=0;
166             int countt=bfs(c1);
167             if(countt==0){
168                 printf("NO\n");
169             }else{
170                 printf("%d\n",countt);
171             }
172         }
173     }
174     return 0;
175 }