HDU 1495 非常可乐(BFS)
题目好理解，汉语的。。。电视上也见过这样的智力游戏，以前看过书，提示是广搜，今天想了一下，思路还是挺清晰的，搜6种情况，如A B C三个杯子，A 到B，A到C，B到A，B到C，C到A,C到B。我实现的很麻烦啊，还好在各种调试，各种DEBUG之后，1Y。
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <string.h> 4 int p[102][102],o1[10000],o2[10000],o3[10000]; 5 int main() 6 { 7 int i,j,s,n,m,start,end,num,z; 8 while(scanf("%d%d%d",&s,&n,&m)!=EOF) 9 { 10 start = end = 1; 11 num = 1; 12 z = 0; 13 memset(p,0,sizeof(p)); 14 if(s == 0&&n == 0&&m == 0) 15 break; 16 o1[1] = s; 17 o2[1] = 0; 18 o3[1] = 0; 19 p[s][0] = 1; 20 while(start<=end) 21 { 22 j = 1; 23 for(i = start; i <= end; i ++) 24 { 25 if(o1[i] != 0) 26 { 27 if(o1[i] >= n - o2[i]) 28 { 29 if(p[o1[i]-n+o2[i]][n]==0) 30 { 31 o1[end+j] = o1[i] - (n - o2[i]); 32 o2[end+j] = n; 33 o3[end+j] = o3[i]; 34 p[o1[i]-n+o2[i]][n] = 1; 35 j ++; 36 } 37 } 38 else 39 { 40 if(p[0][o1[i]+o2[i]]==0) 41 { 42 o1[end+j] = 0; 43 o2[end+j] = o1[i] +o2[i]; 44 o3[end+j] = o3[i]; 45 p[0][o1[i]+o2[i]] = 1; 46 j ++; 47 } 48 } 49 if(o1[i] >= m - o3[i]) 50 { 51 if(p[o1[i]-m+o3[i]][o2[i]]==0) 52 { 53 o1[end+j] = o1[i] - (m - o3[i]); 54 o2[end+j] = o2[i]; 55 o3[end+j] = m; 56 p[o1[i]-m+o3[i]][o2[i]] = 1; 57 j ++; 58 } 59 } 60 else 61 { 62 if(p[0][o2[i]]==0) 63 { 64 o1[end+j] = 0; 65 o2[end+j] = o2[i]; 66 o3[end+j] = o3[i]+o1[i]; 67 p[0][o2[i]] = 1; 68 j ++; 69 } 70 } 71 } 72 if(o2[i] != 0) 73 { 74 if(p[o1[i]+o2[i]][0] == 0) 75 { 76 o1[end+j] = o1[i]+o2[i]; 77 o2[end+j] = 0; 78 o3[end+j] = o3[i]; 79 p[o1[i]+o2[i]][0] = 1; 80 j ++; 81 } 82 if(o2[i] >= m - o3[i]) 83 { 84 if(p[o1[i]][o2[i]-m+o3[i]] == 0) 85 { 86 o1[end+j] = o1[i]; 87 o2[end+j] = o2[i] - m + o3[i]; 88 o3[end+j] = m; 89 p[o1[i]][o2[i]-m+o3[i]] = 1; 90 j ++; 91 } 92 } 93 else 94 { 95 if(p[o1[i]][0]==0) 96 { 97 o1[end+j] = o1[i]; 98 o2[end+j] = 0; 99 o3[end+j] = o2[i] + o3[i]; 100 p[o1[i]][0] = 1; 101 j ++; 102 } 103 } 104 } 105 if(o3[i] != 0) 106 { 107 if(p[o1[i]+o3[i]][o2[i]] == 0) 108 { 109 o1[end+j] = o1[i]+o3[i]; 110 o2[end+j] = o2[i]; 111 o3[end+j] = 0; 112 p[o1[i]+o3[i]][o2[i]] = 1; 113 j ++; 114 } 115 if(o3[i] >= n - o2[i]) 116 { 117 if(p[o1[i]][n] == 0) 118 { 119 o1[end+j] = o1[i]; 120 o2[end+j] = n; 121 o3[end+j] = o2[i] - n + o3[i]; 122 p[o1[i]][n] = 1; 123 j ++; 124 } 125 } 126 else 127 { 128 if(p[o1[i]][o2[i]+o3[i]] == 0) 129 { 130 o1[end+j] = o1[i]; 131 o2[end+j] = o2[i] + o3[i]; 132 o3[end+j] = 0; 133 p[o1[i]][o2[i]+o3[i]] = 1; 134 j ++; 135 } 136 } 137 } 138 } 139 start = end + 1; 140 end = end + j - 1; 141 for(i = start;i <= end;i ++) 142 { 143 if(o1[i] == o3[i]&&o2[i] == 0) 144 { 145 z = 1; 146 break; 147 } 148 if(o2[i] == o3[i]&&o1[i] == 0) 149 { 150 z = 1; 151 break; 152 } 153 if(o1[i] == o2[i]&&o3[i] == 0) 154 { 155 z = 1; 156 break; 157 } 158 159 } 160 if(z) break; 161 num ++; 162 } 163 if(z) 164 printf("%d\n",num); 165 else 166 printf("NO\n"); 167 } 168 return 0; 169 }
