http://poj.org/problem?id=1606
题意：有两个容量为a,b的被子，用这两个被子量出c的水。
思路：bfs和记录路径
1 #include <stdio.h> 2 #include <string.h> 3 #include <queue> 4 5 using namespace std; 6 struct Node{ 7 int pre,a,b; 8 int num,f; 9 }node[2000000]; 10 int pos; 11 queue<Node>s; 12 bool mark[500][500]; 13 14 void print(int t) 15 { 16 switch(t) 17 { 18 case 1: 19 printf("fill A "); 20 break; 21 case 2: 22 printf("fill B "); 23 break; 24 case 3: 25 printf("empty A "); 26 break; 27 case 4: 28 printf("empty B "); 29 break; 30 case 5: 31 printf("pour A B "); 32 break; 33 case 6: 34 printf("pour B A "); 35 break; 36 case 7: 37 printf("success "); 38 default: 39 break; 40 } 41 } 42 void put(int t){ 43 if(t!=0) 44 { 45 put(node[t].pre); 46 print(node[t].f); 47 } 48 } 49 50 void bfs(int a,int b,int c) 51 { 52 while(!s.empty()) 53 s.pop(); 54 node[pos].a = 0; 55 node[pos].b = 0; 56 node[pos].pre = 0; 57 node[pos].num = pos; 58 mark[0][0] = false; 59 s.push(node[pos++]); 60 while(!s.empty()) 61 { 62 Node tmp = s.front(); 63 s.pop(); 64 if(tmp.b==c) 65 { 66 node[pos].f = 7; 67 node[pos].pre = tmp.num; 68 put(pos); 69 return; 70 } 71 if(mark[a][tmp.b]) 72 { 73 node[pos].a = a; 74 node[pos].b = tmp.b; 75 node[pos].num = pos; 76 node[pos].pre = tmp.num; 77 node[pos].f = 1; 78 mark[a][tmp.b] = false; 79 s.push(node[pos++]); 80 } 81 if(mark[tmp.a][b]) 82 { 83 node[pos].a = tmp.a; 84 node[pos].b = b; 85 node[pos].num = pos; 86 node[pos].pre = tmp.num; 87 node[pos].f = 2; 88 mark[tmp.a][b] = false; 89 s.push(node[pos++]); 90 } 91 if(mark[0][tmp.b]) 92 { 93 node[pos].a = 0; 94 node[pos].b = tmp.b; 95 node[pos].num = pos; 96 node[pos].pre = tmp.num; 97 node[pos].f = 3; 98 mark[0][tmp.b] = false; 99 s.push(node[pos++]); 100 } 101 if(mark[tmp.a][0]) 102 { 103 node[pos].a = tmp.a; 104 node[pos].b = 0; 105 node[pos].num = pos; 106 node[pos].pre = tmp.num; 107 node[pos].f = 4; 108 mark[tmp.a][0] = false; 109 s.push(node[pos++]); 110 } 111 if(tmp.a>0) 112 { 113 int t = tmp.a+tmp.b; 114 node[pos].a = t-b<0?0:t-b; 115 node[pos].b = t<b?t:b; 116 if(mark[node[pos].a][node[pos].b]) 117 { 118 node[pos].num = pos; 119 node[pos].pre = tmp.num; 120 node[pos].f = 5; 121 mark[node[pos].a][node[pos].b] = false; 122 s.push(node[pos++]); 123 } 124 } 125 if(tmp.b>0) 126 { 127 int t = tmp.a+tmp.b; 128 node[pos].b = (t-a<0)?0:t-a; 129 node[pos].a = t<a?t:a; 130 if(mark[node[pos].a][node[pos].b]) 131 { 132 node[pos].num = pos; 133 node[pos].pre = tmp.num; 134 node[pos].f = 6; 135 mark[node[pos].a][node[pos].b] = false; 136 s.push(node[pos++]); 137 } 138 } 139 } 140 } 141 142 int main() 143 { 144 int a,b,c; 145 while(~scanf("%d%d%d",&a,&b,&c)) 146 { 147 memset(mark,true,sizeof(mark)); 148 pos = 0; 149 bfs(a,b,c); 150 } 151 return 0; 152 }