题意：就是倒水游戏，有两个杯子无限水可以进行三种操作：
一个杯子装满水，一个杯子倒光，一个杯子倒到另一个杯子。
最后要求其中一个杯子的水量到n，然后输出步骤。
思路:bfs即可，最后递归输出。
代码如下：
1 /************************************************** 2 * Author : xiaohao Z 3 * Blog : http://www.cnblogs.com/shu-xiaohao/ 4 * Last modified : 2014-03-28 18:49 5 * Filename : uva_571.cpp 6 * Description : 7 * ************************************************/ 8 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <cstdlib> 13 #include <cmath> 14 #include <algorithm> 15 #include <queue> 16 #include <stack> 17 #include <vector> 18 #include <set> 19 #include <map> 20 #define MP(a, b) make_pair(a, b) 21 #define PB(a) push_back(a) 22 23 using namespace std; 24 typedef long long ll; 25 typedef pair<int, int> pii; 26 typedef pair<unsigned int,unsigned int> puu; 27 typedef pair<int, double> pid; 28 typedef pair<ll, int> pli; 29 typedef pair<int, ll> pil; 30 31 const int INF = 0x3f3f3f3f; 32 const double eps = 1E-6; 33 const int LEN = 1010; 34 int ca, cb, n; 35 pii vis[LEN][LEN]; 36 37 pii empty(pii x, int a){ 38 if(a == 0)return MP(0, x.second); 39 else return MP(x.first, 0); 40 } 41 42 pii fill(pii x, int a){ 43 if(a == 1) return MP(x.first, cb); 44 else return MP(ca, x.second); 45 } 46 47 pii pour(pii x, int a, int b){ 48 int tmp = x.first + x.second; 49 if(a == 0 && b == 1){ 50 if(tmp <= cb) return MP(0, tmp); 51 else return MP(tmp - cb, cb); 52 }else{ 53 if(tmp <= ca) return MP(tmp, 0); 54 else return MP(ca, tmp - ca); 55 } 56 } 57 58 void flush(pii nvex, pii nv, queue<pii> &q){ 59 if(vis[nv.first][nv.second].first == -1){ 60 vis[nv.first][nv.second] = nvex; 61 q.push(nv); 62 } 63 } 64 65 pii bfs(){ 66 memset(vis, -1, sizeof vis); 67 queue<pii> q; 68 q.push(MP(0, 0)); 69 vis[0][0] = MP(0, 0); 70 while(!q.empty()){ 71 pii nvex = q.front(), nv; q.pop(); 72 if(nvex.first == n || nvex.second == n) return nvex; 73 nv = empty(nvex, 0);flush(nvex, nv, q); 74 nv = empty(nvex, 1);flush(nvex, nv, q); 75 nv = fill(nvex, 0);flush(nvex, nv, q); 76 nv = fill(nvex, 1);flush(nvex, nv, q); 77 nv = pour(nvex, 0, 1);flush(nvex, nv, q); 78 nv = pour(nvex, 1, 0);flush(nvex, nv, q); 79 } 80 return MP(-1, -1); 81 } 82 83 void out(int x, int y){ 84 pii nv, nvex = vis[x][y]; 85 if(vis[x][y] == MP(x, y)) return ; 86 int op; 87 nv = empty(nvex, 0); if(nv == MP(x, y)) op = 1; 88 nv = empty(nvex, 1); if(nv == MP(x, y)) op = 2; 89 nv = fill(nvex, 0); if(nv == MP(x, y)) op = 3; 90 nv = fill(nvex, 1); if(nv == MP(x, y)) op = 4; 91 nv = pour(nvex, 0, 1); if(nv == MP(x, y)) op = 5; 92 nv = pour(nvex, 1, 0); if(nv == MP(x, y)) op = 6; 93 out(vis[x][y].first, vis[x][y].second); 94 switch(op){ 95 case 1: cout << "empty A" << endl;break; 96 case 2: cout << "empty B" << endl;break; 97 case 3: cout << "fill A" << endl;break; 98 case 4: cout << "fill B" << endl;break; 99 case 5: cout << "pour A B" << endl; break; 100 case 6: cout << "pour B A" << endl; break; 101 } 102 } 103 104 int main() 105 { 106 //freopen("in.txt", "r", stdin); 107 108 ios::sync_with_stdio(false); 109 while(cin >> ca >> cb >> n){ 110 pii pos = bfs(); 111 out(pos.first, pos.second); 112 cout << "success" << endl; 113 } 114 return 0; 115 }