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
题意：中文题，不用多说，三个杯子可以互相倒可乐，然而并没有刻度，那么只能依靠杯子的总容量来算了，首先确定当一个杯子有可乐时就可以向其他两个杯子倒可乐了，那么只要判断另外两个杯子的状态就行了，每次像其他杯子倒水都算作一次（可以确定有六种可能的状态了），剩下的就是一个BFS了，就是有点麻烦，代码可长。。。。
#include#include #include #include 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 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;}