这道题做了很长时间，一开始上课的时候手写代码，所以想到了很多细节，但是创客手打代码的时候由于疏忽又未将pair赋初值，导致一直输出错误，以后自己写代码可以专心一点，可能会在宿舍图书馆或者Myhome，创客晚上好吵呀，隔壁真的服...
本题大意：给定两个杯子的容量，有六种操作，通过操作使得两个被子其中一个杯子的水等于待输入值C，并输出最少操作次数和操作名称。
本题思路：BFS，遇到合适的直接输出，搜完都没有搜到就impossible。
参考代码：
将switch改为if能节省大部分篇幅，但是switch更美观明了一点。
1 #include <string> 2 #include <cstring> 3 #include <iostream> 4 #include <queue> 5 #include <stack> 6 using namespace std; 7 8 typedef pair<int ,int > P; 9 const int maxn = 100 + 5, INF = 0x3f3f3f3f; 10 int A, B, C, t; 11 12 struct { 13 int x, y, cnt;//用来存他爹的坐标和倒水的方式 14 } Ans[maxn][maxn]; 15 int vis[maxn][maxn]; 16 string ans[6] = { 17 "FILL(1)", "FILL(2)", "DROP(1)", "DROP(2)", "POUR(1,2)", "POUR(2,1)", 18 }; 19 20 void bfs() { 21 queue <P> Q; 22 vis[0][0] = 0; 23 Ans[0][0].x = -1, Ans[0][0].y = -1; 24 Q.push(make_pair(0, 0)); 25 while(!Q.empty()) { 26 P now = Q.front(), Next; 27 stack <string> s1; 28 Q.pop(); 29 if(now.first == C || now.second == C) { 30 cout << vis[now.first][now.second] << endl; 31 int X = now.first; 32 int Y = now.second; 33 while(X != -1) { 34 s1.push(ans[Ans[X][Y].cnt]); 35 int tmp = X; 36 X = Ans[X][Y].x; 37 Y = Ans[tmp][Y].y; 38 } 39 s1.pop(); 40 while(!s1.empty()) { 41 cout << s1.top() << endl; 42 s1.pop(); 43 } 44 return; 45 } 46 for(int i = 0; i < 6; i ++) { 47 switch(i) { 48 case 0 ://将A加满 49 Next.first = A; 50 Next.second = now.second; 51 break; 52 case 1 ://将B加满 53 Next.second = B; 54 Next.first = now.first; 55 break; 56 case 2 ://将A倒掉 57 Next.second = now.second; 58 Next.first = 0; 59 break; 60 case 3 ://将B倒掉 61 Next.first = now.first; 62 Next.second = 0; 63 break; 64 case 4 ://将A倒入B中 65 t = B - now.second;//B中还能倒入的量 66 Next.first = now.first - t; 67 if(Next.first < 0) Next.first = 0; 68 Next.second = now.first + now.second; 69 if(Next.second > B) Next.second = B; 70 break; 71 case 5 ://将B倒入A中 72 t = A - now.first;//A中还能倒入的量 73 Next.second = now.second - t; 74 if(Next.second < 0) Next.second = 0; 75 Next.first = now.first + now.second; 76 if(Next.first > A) Next.first = A; 77 break; 78 } 79 if(vis[Next.first][Next.second] == INF) { 80 vis[Next.first][Next.second] = vis[now.first][now.second] + 1; 81 Q.push(Next); 82 Ans[Next.first][Next.second].x = now.first; 83 Ans[Next.first][Next.second].y = now.second; 84 Ans[Next.first][Next.second].cnt = i; 85 } 86 } 87 } 88 cout << "impossible" << endl; 89 } 90 91 int main () { 92 memset(vis, INF, sizeof(vis)); 93 cin >> A >> B >> C; 94 bfs(); 95 return 0; 96 }