题目：http://poj.org/problem?id=3414
题意：给出了两个瓶子的容量A,B, 以及一个目标水量C，
对A、B可以有如下操作:
FILL(i) fill the pot i (1 ≤ i ≤ 2) from the tap;
DROP(i) empty the pot i to the drain;
POUR(i,j) pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j).
问经过哪几个操作后能使得任意一个瓶子的残余水量为C。
若不可能得到则输出impossible
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<stack> 6 #include<queue> 7 #include<iomanip> 8 #include<cmath> 9 #include<map> 10 #include<vector> 11 #include<algorithm> 12 using namespace std; 13 14 int vis[110][110]; 15 int a,b,c; 16 struct node 17 { 18 int x,y,step; 19 }pos,next; 20 21 struct way 22 { 23 int x,y,f; 24 }before[110][110]; 25 26 void pri(int x,int y) 27 { 28 stack<int>st; 29 while(x!=0||y!=0) 30 { 31 st.push(before[x][y].f); 32 int tx=before[x][y].x; 33 int ty=before[x][y].y; 34 x=tx; y=ty; 35 } 36 while(!st.empty()) 37 { 38 switch(st.top()) 39 { 40 case 1:printf("DROP(1) "); break; 41 case 2:printf("DROP(2) "); break; 42 case 3:printf("FILL(1) "); break; 43 case 4:printf("FILL(2) "); break; 44 case 5:printf("POUR(1,2) "); break; 45 case 6:printf("POUR(2,1) "); break; 46 } 47 st.pop(); 48 } 49 } 50 int bfs() 51 { 52 queue<node>q; 53 next.x=0; next.y=0; next.step=0; 54 vis[0][0]=1; 55 q.push(next); 56 while(!q.empty()) 57 { 58 pos=q.front(); 59 //cout<<pos.x<<" "<<pos.y<<" "<<pos.step<<" "; 60 //cout<<before[pos.x][pos.y].x<<" "<<before[pos.x][pos.y].y<<" "<<before[pos.x][pos.y].f<<endl; 61 q.pop(); 62 if(pos.x==c||pos.y==c) 63 { 64 cout<<pos.step<<endl; 65 pri(pos.x,pos.y); 66 return 1; 67 } 68 if(!vis[0][pos.y]&&pos.x!=0) 69 { 70 next.x=0; next.y=pos.y; next.step=pos.step+1; 71 q.push(next); 72 vis[0][pos.y]=1; 73 before[0][pos.y]=(struct way){pos.x,pos.y,1}; 74 } 75 if(!vis[pos.x][0]&&pos.y!=0) 76 { 77 next.x=pos.x; next.y=0; next.step=pos.step+1; 78 q.push(next); 79 vis[pos.x][0]=1; 80 before[pos.x][0]=(struct way){pos.x,pos.y,2}; 81 } 82 if(!vis[a][pos.y]&&pos.x!=a) 83 { 84 next.x=a; next.y=pos.y; next.step=pos.step+1; 85 q.push(next); 86 vis[a][pos.y]=1; 87 before[a][pos.y]=(struct way){pos.x,pos.y,3}; 88 } 89 if(!vis[pos.x][b]&&pos.y!=b) 90 { 91 next.x=pos.x; next.y=b; next.step=pos.step+1; 92 q.push(next); 93 vis[pos.x][b]=1; 94 before[pos.x][b]=(struct way){pos.x,pos.y,4}; 95 } 96 if(pos.x>0&&pos.y<b) 97 { 98 int t=min(pos.x,b-pos.y); 99 if(!vis[pos.x-t][pos.y+t]) 100 { 101 q.push((struct node){pos.x-t,pos.y+t,pos.step+1}); 102 vis[pos.x-t][pos.y+t]=1; 103 before[pos.x-t][pos.y+t]=(struct way){pos.x,pos.y,5}; 104 } 105 } 106 if(pos.x<a&&pos.y>0) 107 { 108 int t=min(pos.y,a-pos.x); 109 if(!vis[pos.x+t][pos.y-t]) 110 { 111 q.push((struct node){pos.x+t,pos.y-t,pos.step+1}); 112 vis[pos.x+t][pos.y-t]=1; 113 before[pos.x+t][pos.y-t]=(struct way){pos.x,pos.y,6}; 114 } 115 } 116 } 117 return 0; 118 } 119 int main() 120 { 121 cin>>a>>b>>c; 122 memset(vis,0,sizeof(vis)); 123 if(bfs()==0) 124 cout<<"impossible"<<endl; 125 return 0; 126 } 127