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Problem Description
大家一定觉的运动以后喝可乐是一件很惬意的事情，但是seeyou却不这么认为。因为每次当seeyou买了可乐以后，阿牛就要求和seeyou一起分享 这一瓶可乐，而且一定要喝的和seeyou一样多。但seeyou的手中只有两个杯子，它们的容量分别是N 毫升和M 毫升 可乐的体积为S （S<101）毫升 (正好装满一瓶) ，它们三个之间可以相互倒可乐 (都是没有刻度的，且 S==N+M，101＞S＞0，N＞0，M＞0) 。聪明的ACMER你们说他们能平分吗？如果能请输出倒可乐的最少的次数，如果不能输出"NO"。
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Input
三个整数 : S 可乐的体积 , N 和 M是两个杯子的容量，以"0 0 0"结束。
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Output
如果能平分的话请输出最少要倒的次数，否则输出"NO"。
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Sample Input
7 4 3 4 1 3 0 0 0 |
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Sample Output
NO 3 |
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Code  View Code
1 #include <stdio.h> 2 #include <queue> 3 using namespace std; 4 struct State 5 { 6 int v[3]; 7 int step; //because this step of pre and cur are maybe the same 8 }; 9 int volume[3]; 10 int BFS() 11 { 12 int map[105][105] = {0}; //initialize and for mark 13 queue<State> q; 14 State cur, pre; 15 int i, j; 16 //initialize the node and push it into the queue 17 cur.v[0] = volume[0]; 18 cur.v[1] = 0; 19 cur.v[2] = 0; 20 cur.step = 0; 21 q.push(cur); 22 map[0][0] = 1; 23 while (!q.empty()) 24 { 25 pre = q.front(); 26 q.pop(); 27 //printf("\nstep = %d\n", pre.step); 28 //printf("pre: %d %d %d\n", pre.v[0], pre.v[1], pre.v[2]); 29 /* 30 v[i] -> v[j], when v[i] have water 31 we should consider that when u pull water from v[i] to v[j], v[j] maybe will arive the volume 32 */ 33 for (i = 0; i < 3 && pre.v[i] > 0; i++) 34 { 35 for (j = 0; j < 3; j++) 36 { 37 if (i != j) 38 { 39 cur = pre; 40 cur.step = pre.step + 1; 41 //when v[j] arrive its volume 42 if (cur.v[i] + cur.v[j] < volume[j]) 43 { 44 cur.v[j] = cur.v[i] + cur.v[j]; 45 cur.v[i] = 0; 46 } 47 else 48 { 49 cur.v[i] = cur.v[i] - (volume[j] - cur.v[j]); 50 cur.v[j] = volume[j]; 51 } 52 //judge whether u have traversed the state 53 if (map[cur.v[1]][cur.v[2]] == 0) 54 { 55 if (cur.v[0] == volume[0] / 2 && cur.v[1] == volume[0] / 2) 56 return cur.step; 57 q.push(cur); 58 //printf("cur: %d %d %d\n", cur.v[0], cur.v[1], cur.v[2]); 59 map[cur.v[1]][cur.v[2]] = 1; 60 } 61 } 62 } 63 } 64 } 65 return 0; 66 } 67 int main() 68 { 69 int result; 70 while (scanf("%d %d %d", &volume[0], &volume[1], &volume[2]) != EOF && (volume[0] || volume[1] || volume[2])) 71 { 72 if (volume[1] < volume[2]) 73 { 74 int temp = volume[1]; 75 volume[1] = volume[2]; 76 volume[2] = temp; 77 } 78 result = BFS(); 79 if (volume[0] % 2 != 0) 80 puts("NO"); 81 else if (result == 0) 82 puts("NO"); 83 else 84 printf("%d\n", BFS()); 85 } 86 return 0; 87 } |
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Idea Firstly, I use the "辗转相除", it's no doubt that I am wrong. We can use BFS to solve this question. Well, this is still other question that it's hard to describe the translation of state. Maybe u can write one by one, but I find that we can use the struct including v[3] to solve this complicated question. Well, next pic show the every step.
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Recommend
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