分油问题

来源：互联网发布：白之王面具淘宝编辑：程序博客网时间：2024/06/02 12:32

递归

状态（局面）---图中的点

合法操作---图中的边

转化为求节点通路问题。

注意：节点动态产生

思路：不断产生新状态，直到不能产生新的状态为止，在这个过程中如果碰到了要求的某个状态，则有解，否则无解。

package bishi;import java.util.HashSet;import java.util.Set;class OilStatus {static int[] full = { 12, 8, 5 };// 3个油桶最大容量int[] v = new int[3];// 3个油桶装油量OilStatus from;// 从哪个状态来public OilStatus(int a, int b, int c) {v[0] = a;v[1] = b;v[2] = c;}public Set<OilStatus> op() {// 返回当前状态可以生成的状态集合Set<OilStatus> rt = new HashSet<OilStatus>();for (int i = 0; i < v.length; i++) {// 笛卡尔积for (int j = 0; j < v.length; j++) {if (i == j) {// 倒入自身continue;}if (v[i] == 0) {// 原容器为空continue;}if (v[j] == full[j]) {// 目标容器已满continue;}OilStatus ost = new OilStatus(v[0], v[1], v[2]);ost.from = this;ost.v[j] += ost.v[i];ost.v[i] = 0;if (ost.v[j] > full[j]) {// 溢出ost.v[i] = ost.v[j] - full[j];ost.v[j] = full[j];}rt.add(ost);}}return rt;}public String toString() {return "<" + v[0] + "," + v[1] + "," + v[2] + ">";}public int hashCode() {return 100;}public boolean equals(Object x) {OilStatus obj = (OilStatus) x;return obj.v[0] == this.v[0] && obj.v[1] == this.v[1]&& obj.v[2] == obj.v[2];}public boolean has(int x) {return this.v[0] == x || this.v[1] == x || this.v[2] == x;}public OilStatus getFrom() {// 返回当前状态的上一状态return this.from;}}public class OilDivision {public static void main(String[] args) {Set<OilStatus> st = new HashSet<OilStatus>();// 存放所有状态st.add(new OilStatus(12, 0, 0));// 初始状态为(12，0，0)for (;;) {Set<OilStatus> new_st = new HashSet<OilStatus>();// 新生成的状态for (Object obj : st) {OilStatus os = (OilStatus) obj;Set<OilStatus> tmp = os.op();// 生成新状态new_st.addAll(tmp);// 添加所有的新状态}if (st.containsAll(new_st)) {// containsAll新生成的状态都在原有状态中break;}st.addAll(new_st);// addAll，加入到原有状态中，集合求并集}for (OilStatus k : st) {OilStatus ost = (OilStatus) k;if (ost.has(6)) {// 目标油量为6while (ost != null) {// 打印状态变换过程System.out.println(ost);ost = ost.getFrom();}break;}}}}

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