这道题蛮有意思的，大意是有三个水桶ABC，只有C水桶放满水，从一个水桶倒到另一个水桶，要么一个桶空，要么一个桶满为止，问说当A桶空时，C桶可能出现的所有的可能状态。
比较容易想到的是讲(a,b,c)表示为状态，然后暴力搜索出所有的可能性，因为一共就3个桶，操作也不复杂，对使用过的状态进行标记，这样可以在O(N3)完成，因为USACO的数据很小,<=20,所以就直接AC掉了，看了ANALYSIS之后知道，其实状态用N2就可以搞定，这样O(N2)就可以完成了。
1 #include <iostream>
2 #include <string>
3 #include <algorithm>
4 #include <string.h>
5 #include <vector>
6 #include <math.h>
7 #include <time.h>
8 #include <queue>
9 using namespace std;
10
11 const int MAX = 1005;
12
13 struct WATER
14 {
15 int a, c;
16 WATER(int aa, int cc) : a(aa), c(cc) {}
17 WATER(){}
18 };
19
20 int A, B, C;
21 queue<WATER>water;
22 vector<int>ans;
23 bool use[MAX][MAX];
24 //存a和c的水量就行b = C - a - c
25 //状态只有O(n*n)
26
27 void pour(int& x, int& y, int X, int Y)
28 {
29 if(x <= Y - y)
30 {
31 y += x;
32 x = 0;
33 }
34 else
35 {
36 x -= (Y - y);
37 y = Y;
38 }
39 }
40
41 void judge(int a, int b, int c)
42 {
43 if(use[a][c] == false)
44 {
45 use[a][c] = true;
46 water.push(WATER(a, c));
47 }
48 }
49
50 void check(int a, int b, int c)
51 {
52 int aa = a;
53 int bb = b;
54 int cc = c;
55
56 pour(a, b, A, B);
57 judge(a, b, c);
58 a = aa, b = bb, c = cc;
59
60 pour(a, c, A, C);
61 judge(a, b, c);
62 a = aa, b = bb, c = cc;
63
64 pour(b, a, B, A);
65 judge(a, b, c);
66 a = aa, b = bb, c = cc;
67
68 pour(b, c, B, C);
69 judge(a, b, c);
70 a = aa, b = bb, c = cc;
71
72 pour(c, a, C, A);
73 judge(a, b, c);
74 a = aa, b = bb, c = cc;
75
76 pour(c, b, C, B);
77 judge(a, b, c);
78 a = aa, b = bb, c = cc;
79 }
80
81 void go()
82 {
83 while(!water.empty())
84 {
85 WATER w = water.front();
86 if(w.a == 0) ans.push_back(w.c);
87 water.pop();
88 check(w.a, C - w.a - w.c, w.c);
89 }
90
91 sort(ans.begin(), ans.end());
92 for(int i = 0; i < ans.size(); i++)
93 {
94 if(i) printf(" %d", ans[i]);
95 else printf("%d", ans[i]);
96 }
97 printf("\n");
98 }
99
100 int main()
101 {
102 freopen("milk3.in", "r", stdin);
103 freopen("milk3.out", "w", stdout);
104
105 int a, b, c;
106 scanf("%d%d%d", &A, &B, &C);
107 water.push(WATER(0, C));
108 memset(use, 0, sizeof(use));
109 use[0][C] = true;
110 go();
111 }