USACO 1.4 Mother's Milk

最新推荐文章于 2021-11-04 22:01:55 发布

转载最新推荐文章于 2021-11-04 22:01:55 发布 · 58 阅读

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原文链接：http://www.cnblogs.com/superbin/archive/2010/05/22/1741634.html

本文探讨了通过BFS算法解决三个杯子互相倒水的问题，目标是在特定条件下找到A空杯时C杯中的牛奶量序列。

比较经典的3个杯子互相倒水的问题。

状态即3个杯子中的水量，用一个3维数组记录，转移为六种组合。

code

1 /*
2 ID: superbi1
3 LANG: C
4 TASK: milk3
5 */
6 //BFS
7 //三个杯子A,B,C 刚开始，C中milk满的，然后根据规则倒，问A为空时的C的milk量的序列
8 #include <stdio.h>
9 #include <string.h>
10 #include <stdlib.h>
11
12 //其实可以用三维数组记录状态
13 char flg[1000000];
14
15 main()
16 {
17 FILE *fin = fopen("milk3.in", "r");
18 FILE *fout = fopen("milk3.out", "w");
19 int a, b, c;
20 int milk[21];
21 int start = c;
22 int que[1000], top = 0, mk;
23 int cur;
24 int I, a0, b0, c0, a1, b1, c1;
25 fscanf(fin, "%d%d%d", &a, &b, &c);
26 memset(flg, 0, sizeof(flg));
27 que[top++] = c;
28 flg[c] = 1;
29 for (I=0; I<21; I++) milk[I] = 0;
30 while (top > 0) {
31 cur = que[--top];
32 if (!(cur/10000)) {
33 milk[cur%100] = 1;
34 }
35 a0 = cur/10000;
36 b0 = (cur%10000)/100;
37 c0 = cur%100;
38 a1 = a0;
39 b1 = b0;
40 c1 = c0;
41 //枚举六种倒的情况
42 if (a1 > 0) {
43 if (b1 < b) {
44 if (b-b1 < a1) {
45 a1 -= b-b1;
46 b1 = b;
47 }else {
48 b1 += a1;
49 a1 = 0;
50 }
51 if (!flg[a1*10000+b1*100+c1]) {
52 flg[a1*10000+b1*100+c1] = 1;
53 que[top++] = a1*10000+b1*100+c1;
54 }
55 }
56 a1 = a0;
57 b1 = b0;
58 c1 = c0;
59 if (c1 < c) {
60 if (c-c1 < a1) {
61 a1 -= c-c1;
62 c1 = c;
63 }else {
64 c1 += a1;
65 a1 = 0;
66 }
67 if (!flg[a1*10000+b1*100+c1]) {
68 flg[a1*10000+b1*100+c1] = 1;
69 que[top++] = a1*10000+b1*100+c1;
70 }
71 }
72 }
73
74 a1 = a0;
75 b1 = b0;
76 c1 = c0;
77 if (b1 > 0) {
78 if (a1 < a) {
79 if (a-a1 < b1) {
80 b1 -= a-a1;
81 a1 = a;
82 }else {
83 a1 += b1;
84 b1 = 0;
85 }
86 if (!flg[a1*10000+b1*100+c1]) {
87 flg[a1*10000+b1*100+c1] = 1;
88 que[top++] = a1*10000+b1*100+c1;
89 }
90 }
91 a1 = a0;
92 b1 = b0;
93 c1 = c0;
94 if (c1 < c) {
95 if (c-c1 < b1) {
96 b1 -= c-c1;
97 c1 = c;
98 }else {
99 c1 += b1;
100 b1 = 0;
101 }
102 if (!flg[a1*10000+b1*100+c1]) {
103 flg[a1*10000+b1*100+c1] = 1;
104 que[top++] = a1*10000+b1*100+c1;
105 }
106 }
107 }
108
109 a1 = a0;
110 b1 = b0;
111 c1 = c0;
112 if (c1 > 0) {
113 if (a1 < a) {
114 if (a-a1 < c1) {
115 c1 -= a-a1;
116 a1 = a;
117 }else {
118 a1 += c1;
119 c1 = 0;
120 }
121 if (!flg[a1*10000+b1*100+c1]) {
122 flg[a1*10000+b1*100+c1] = 1;
123 que[top++] = a1*10000+b1*100+c1;
124 }
125 }
126 a1 = a0;
127 b1 = b0;
128 c1 = c0;
129 if (b1 < b) {
130 if (b-b1 < c1) {
131 c1 -= b-b1;
132 b1 = b;
133 }else {
134 b1 += c1;
135 c1 = 0;
136 }
137 if (!flg[a1*10000+b1*100+c1]) {
138 flg[a1*10000+b1*100+c1] = 1;
139 que[top++] = a1*10000+b1*100+c1;
140 }
141 }
142 }
143 }
144 mk = 0;
145 for (I=0; I<21; I++) {
146 if (milk[I]) {
147 if (!mk) { fprintf(fout, "%d", I); mk = 1; }
148 else fprintf(fout, " %d", I);
149 }
150 }
151 fprintf(fout, "\n");
152 exit(0);
153 }

转载于:https://www.cnblogs.com/superbin/archive/2010/05/22/1741634.html