7 4 3
4 1 3
0 0 0

NO
3分析：这道题最终的状态就是三个里面只要有两个体积相等同时另一个为零即可，那么由于可能性很多并且步数繁杂，采用广搜是比较好的方法，虽然代码量大，但是基本思路都是重复的，大致思路就是把六种状态都遍历一遍就好了

import java.util.*;public class Main {static Scanner in = new Scanner(System.in);static final int maxn = 105;//标志数组来标记路径，三个杯子三维数组即可static int[][][] visited = new int[maxn][maxn][maxn];static void init() {for (int i = 0; i < maxn; ++i)for (int j = 0; j < maxn; ++j)for (int k = 0; k < maxn; ++k)visited[i][j][k] = 0;}// a b c为三个容器的最大容量static void bfs(int a, int b, int c) {if (a % 2 == 1) {System.out.println("NO");return;}Queue<Node> que = new LinkedList<Node>();// 初始que.add(new Node(a, 0, 0, 0));while (!que.isEmpty()) {// 取队头并弹出Node t = que.poll();visited[t.a][t.b][t.c] = 1;// 判断是否符合条件if (t.a == t.b && t.c == 0 || t.a == t.c && t.b == 0 || t.b == t.c && t.a == 0) {System.out.println(t.step);return;}// 倒水过程，注意倒水的前提是杯子里面有水// b -> aif (t.b != 0) {// 因为没有刻度， 所以每次倒水都有两种情况// 第一种情况是把自己倒完if (t.a + t.b <= a) {if (visited[t.a + t.b][0][t.c] == 0) {que.add(new Node(t.a + t.b, 0, t.c, t.step + 1));visited[t.a + t.b][0][t.c] = 1;}}// 第二种情况是把对方倒满else if (t.a != a) {if (visited[a][t.b - (a - t.a)][t.c] == 0) {que.add(new Node(a, t.b - (a - t.a), t.c, t.step + 1));visited[a][t.b - (a - t.a)][t.c] = 1;}}}// c -> aif (t.c != 0) {if (t.a + t.c <= a) {if (visited[t.a + t.c][t.b][0] == 0) {que.add(new Node(t.a + t.c, t.b, 0, t.step + 1));visited[t.a + t.c][t.b][0] = 1;}} else if (t.a != a) {if (visited[a][t.b][t.c - (a - t.a)] == 0) {que.add(new Node(a, t.b, t.c - (a - t.a), t.step + 1));visited[a][t.b][t.c - (a - t.a)] = 1;}}}// b -> cif (t.b != 0) {if (t.b + t.c <= c) {if (visited[t.a][0][t.b + t.c] == 0) {que.add(new Node(t.a, 0, t.b + t.c, t.step + 1));visited[t.a][0][t.b + t.c] = 1;}} else if (t.c != c) {if (visited[t.a][t.b - (c - t.c)][c] == 0) {que.add(new Node(t.a, t.b - (c - t.c), c, t.step + 1));visited[t.a][t.b - (c - t.c)][c] = 1;}}}// c -> bif (t.c != 0) {if (t.c + t.b <= b) {if (visited[t.a][t.c + t.b][0] == 0) {que.add(new Node(t.a, t.c + t.b, 0, t.step + 1));visited[t.a][t.c + t.b][0] = 1;}} else if (t.b != b) {if (visited[t.a][b][t.c - (b - t.b)] == 0) {que.add(new Node(t.a, b, t.c - (b - t.b), t.step + 1));visited[t.a][b][t.c - (b - t.b)] = 1;}}}// a -> bif (t.a != 0) {if (t.a + t.b <= b) {if (visited[0][t.a + t.b][t.c] == 0) {que.add(new Node(0, t.a + t.b, t.c, t.step + 1));visited[0][t.a + t.b][t.c] = 1;}} else if (t.b != b) {if (visited[t.a - (b - t.b)][b][t.c] == 0) {que.add(new Node(t.a - (b - t.b), b, t.c, t.step + 1));visited[t.a - (b - t.b)][b][t.c] = 1;}}}// a -> cif (t.a != 0) {if (t.a + t.c <= c) {if (visited[0][t.b][t.a + t.c] == 0) {que.add(new Node(0, t.b, t.a + t.c, t.step + 1));visited[0][t.b][t.a + t.c] = 1;}} else if (t.c != c) {if (visited[t.a - (c - t.c)][t.b][c] == 0) {que.add(new Node(t.a - (c - t.c), t.b, c, t.step + 1));visited[t.a - (c - t.c)][t.b][c] = 1;}}}}System.out.println("NO");}public static void main(String[] args) {while (in.hasNext()) {int a = in.nextInt(), b = in.nextInt(), c = in.nextInt();if (a == 0 && b == 0 && c == 0)break;init();bfs(a, b, c);}}}class Node {// a b c 代表实际拥有水的体积int a, b, c, step;Node(int a, int b, int c, int step) {this.a = a;this.b = b;this.c = c;this.step = step;}
}