http://poj.org/problem?id=3414
题意：给出1号杯子的容量a，2号杯子的容量b，目标盛水量c。问两个杯子经过怎样的变换能盛出c容量的水，输出最小步数及转换过程，如不能转换，输出“‘impossible”。
思路：总共有6种转换，bfs搜索每一种转换，并记录前一个状态。最后将记录的状态倒着输出。
1 #include <stdio.h> 2 #include <iostream> 3 #include <string.h> 4 #include <queue> 5 #include <stack> 6 const int N=112; 7 using namespace std; 8 string oper[7] = {"","FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(2,1)","POUR(1,2)"}; 9 //六种操作 10 int a,b,c,vis[N][N],keep[N];//vis[][]用来标记当前状态，keep[][]用来存储记录的id 11 struct node 12 { 13 int x; 14 int y; 15 int step; 16 }; 17 struct path 18 { 19 int x; 20 int y; 21 int id; 22 } pre[N][N];//记录当前状态下的前一个状态 23 void show(int x,int y) 24 { 25 int k = 0; 26 while (x||y) 27 { 28 int xx = pre[x][y].x; 29 int yy = pre[x][y].y; 30 int id = pre[x][y].id; 31 x = xx; 32 y = yy; 33 keep[k++] = id;//存储达到目标存水量c的之前的操作id 34 } 35 for (int i = k-1; i >= 0; i --) 36 cout<<oper[keep[i]]<<endl; 37 38 } 39 void bfs() 40 { 41 queue<node>q; 42 q.push((struct node) 43 { 44 0,0,0 45 }); 46 vis[0][0] = 1; 47 while(!q.empty()) 48 { 49 node t = q.front(); 50 q.pop(); 51 if (t.x==c||t.y==c) 52 { 53 printf("%d ",t.step); 54 show(t.x,t.y); 55 return ; 56 } 57 if(t.x < a && !vis[a][t.y])//FILL(1) 58 { 59 q.push((struct node){a,t.y,t.step+1}); 60 vis[a][t.y] = 1; 61 pre[a][t.y] = (struct path){t.x,t.y,1}; 62 } 63 if (t.y < b && !vis[t.x][b])//FILL(2) 64 { 65 q.push((struct node){t.x,b,t.step+1}); 66 vis[t.x][b] = 1; 67 pre[t.x][b] = (struct path){t.x,t.y,2}; 68 } 69 if (t.x > 0 && !vis[0][t.y])//DROP(1) 70 { 71 q.push((struct node){0,t.y,t.step+1}); 72 vis[0][t.y] = 1; 73 pre[0][t.y] = (struct path){t.x,t.y,3}; 74 } 75 if (t.y > 0 && !vis[t.x][0])//DROP(2) 76 { 77 q.push((struct node){t.x,0,t.step+1}); 78 vis[t.x][0] = 1; 79 pre[t.x][0] = (struct path){t.x,t.y,4}; 80 } 81 if (t.y > 0 && t.x < a)//POUR(2,1) 82 { 83 int m = min(a-t.x,t.y); 84 if (!vis[t.x+m][t.y-m]) 85 { 86 q.push((struct node){t.x+m,t.y-m,t.step+1}); 87 vis[t.x+m][t.y-m] = 1; 88 pre[t.x+m][t.y-m] = (struct path){t.x,t.y,5}; 89 } 90 } 91 if (t.x > 0 && t.y < b)//POUR(1,2) 92 { 93 int m = min(b-t.y,t.x); 94 if (!vis[t.x-m][t.y+m]) 95 { 96 q.push((struct node){t.x-m,t.y+m,t.step+1}); 97 vis[t.x-m][t.y+m] = 1; 98 pre[t.x-m][t.y+m] = (struct path){t.x,t.y,6}; 99 } 100 } 101 } 102 printf("impossible "); 103 } 104 int main() 105 { 106 scanf("%d%d%d",&a,&b,&c); 107 memset(vis,0,sizeof(vis)); 108 bfs(); 109 return 0; 110 }