HDU 1495 非常可乐-BFS-[解题报告] C++ | Acm之家


HDU 1495 非常可乐-BFS-[解题报告] C++ | Acm之家
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"。
输入:
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。
输出:
如果能平分的话请输出最少要倒的次数,否则输出"NO"。
样例输入:
7 4 3 4 1 3 0 0 0
样例输出:
NO 3


思路:明确目标状态是有两个杯子都是S/2,那么,只要找到这状态就行,可能从开始状态出发,把每一种出现的状态都找出来,并保存,直到找到这状态.当然,重复状态要过虑

06using namespace std;
07const int maxn=100+5;
08bool hash[maxn][maxn];
09 
10int a[5];
11struct node {
12    int v[3],step;
13};
14void BFS() {
15    node cur,next;
16    cur.v[0]=a[0];
17    cur.v[1]=cur.v[2]=0;
18    cur.step=0;
19    queue<node>q;
20    memset(hash,0,sizeof(hash));
21    hash[0][0]=1;
22    q.push(cur);
23    while(!q.empty()) {
24        cur=q.front();
25        q.pop();
26        if((cur.v[0]==a[0]/2&&cur.v[1]==a[0]/2)
27           ||(cur.v[0]==a[0]/2&&cur.v[2]==a[0]/2)
28           ||(cur.v[1]==a[0]/2&&cur.v[2]==a[0]/2)){
29            printf("%d\n",cur.step); return ;///出现了两个杯子各占一半,找到状态
30        }
31        for(int i=0; i<3; i++)///找到还有水的杯子
32            if(cur.v[i]>0)
33                for(int j=0; j<3; j++) {
34                    next=cur;
35                    if(i==j) continue;///不能倒给自已
36                    if(next.v[i]+next.v[j]>a[j]) {
37                        next.v[i]-=a[j]-next.v[j];///装满另一个杯子还有多
38                        next.v[j]=a[j];///杯子满了
39                    }
40                    else {
41                        next.v[j]+=next.v[i];///还没满或刚好满
42                        next.v[i]=0;///全倒完
43                    }
44                    if(!hash[next.v[1]][next.v[2]]) {///这种状态没有出现过
45                        hash[next.v[1]][next.v[2]]=1;
46                        next.step++;
47                        q.push(next);
48                    }
49                }
50    }
51    printf("NO\n");
52}
53 
54int main() {
55    while(~scanf("%d%d%d",&a[0],&a[1],&a[2]),a[0]||a[1]||a[2]) {
56        if(a[0]%2) {///因为三个杯子容量都是整数
57            puts("NO"); continue;
58        }
59        if(a[1]==a[2]) {///刚好平分
60            puts("1"); continue;
61        }
62        BFS();
63    }
64    return 0;
65}
http://java-mans.iteye.com/blog/1645092

http://blog.csdn.net/deng_hui_long/article/details/9957439
Read full article from HDU 1495 非常可乐-BFS-[解题报告] C++ | Acm之家

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