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1 /*
2 题意:给了两堆牌,每次从首部取出一张牌,按颜色分配到两个新堆,分配过程两新堆的总数差不大于1
3 记忆化搜索(DFS+DP):我们思考如果我们将连续的两个操作看成一个集体操作,那么这个操作必然是1红1黑
4 考虑三种情况:a[]连续两个颜色相同,输出11;b[]连续两个相同,输出22;
5 a[x] != b[y], 输出12;否则Impossible
6 详细解释:http://blog.csdn.net/jsun_moon/article/details/10254417
7 */
8 #include <cstdio>
9 #include <iostream>
10 #include <algorithm>
11 #include <cstring>
12 #include <cmath>
13 #include <string>
14 using namespace std;
15
16 const int MAXN = 1e3 + 10;
17 const int INF = 0x3f3f3f3f;
18 char a[MAXN], b[MAXN];
19 bool dp[MAXN][MAXN];
20
21 bool DFS(int x, int y)
22 {
23 if (!x && !y) return true;
24 if (!dp[x][y]) dp[x][y] = true;
25 else return false;
26
27 if ((x-2)>=0 && a[x] != a[x-1] && DFS (x-2, y))
28 {
29 printf ("11"); return true;
30 }
31 if ((y-2)>=0 && b[y] != b[y-1] && DFS (x, y-2))
32 {
33 printf ("22"); return true;
34 }
35 if ((x-1)>=0 && (y-1)>=0 && a[x] != b[y] && DFS (x-1, y-1))
36 {
37 printf ("12"); return true;
38 }
39
40 return false;
41 }
42
43 int main(void) //URAL 1501 Sense of Beauty
44 {
45 //freopen ("V.in", "r", stdin);
46
47 int n;
48 while (scanf ("%d", &n) == 1)
49 {
50 memset (dp, false, sizeof (dp));
51 scanf ("%s", a+1); scanf ("%s", b+1);
52
53 if (!DFS (n, n)) puts ("Impossible");
54 else puts ("");
55 }
56
57 return 0;
58 }
59
60 /*
61 Impossible
62 */
记忆化搜索(DFS+DP) URAL 1501 Sense of Beauty
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原文地址:http://www.cnblogs.com/Running-Time/p/4495645.html