UVA 10651 Pebble Solitaire 状态压缩dp

一开始还在纠结怎么表示一个状态，毕竟是一个串。后来搜了一下题解发现了这里用一个整数的前12位表示转态就好了 ，1~o，0~'-'，每个状态用一个数来表示，然后dp写起来就比较方便了。

代码：

 1 #include <iostream>
 2 #include <sstream>
 3 #include <cstdio>
 4 #include <climits>
 5 #include <cstring>
 6 #include <cstdlib>
 7 #include <string>
 8 #include <cmath>
 9 #include <vector>
10 #include <queue>
11 #include <algorithm>
12 #define esp 1e-6
13 #define pb push_back
14 #define in  freopen("in.txt", "r", stdin);
15 #define out freopen("out.txt", "w", stdout);
16 #define print(a) printf("%d\n",(a));
17 #define bug puts("********))))))");
18 #define Rep(i, c) for(__typeof(c.end()) i = c.begin(); i != c.end(); i++)
19 #define inf 0x0f0f0f0f
20 using namespace std;
21 typedef long long  LL;
22 typedef vector<int> VI;
23 typedef pair<int, int> pii;
24 typedef vector<pii,int> VII;
25 typedef vector<int>:: iterator IT;
26 #define N 50000
27 int dp[N];
28 int ans;
29 void f(int x)
30 {
31     if(dp[x])
32         return ;
33     int num = 0;
34     for(int i = 0; i < 12; i++)
35         if(x&(1<<i))
36         num++;
37     ans = min(ans, num);
38     dp[x] = 1;
39     for(int i = 0; i <= 9; i++)
40         if(((x&(1<<i)) && (x&(1<<(i+1))) && !(x&(1<<(i+2))))
41                 ||(!(x&(1<<i)) && (x&(1<<(i+1))) && (x&(1<<(i+2)))))
42             f(x^(1<<(i+1))^(1<<i)^(1<<(i+2)));
43 }
44 int main(void)
45 {
46 
47     char s[100];
48     int T, t;
49     for(t = scanf("%d", &T), gets(s); t <= T; t++)
50     {
51         memset(dp, 0, sizeof(dp));
52         gets(s);
53         int n = 0;
54         ans = 20;
55         for(int i = 0; i < 12; i++)
56             if(s[i] - '-')
57                 n ^= (1<<i);
58                 f(n);
59         printf("%d\n", ans);
60     }
61     return 0;
62 }

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或者可以用map+string的方法，一直都没怎么学过STL,看了http://www.myexception.cn/ai/1243266.html 这里的方法表示又涨了不少知识，拿来重新写了一遍（其实还是“剽窃”，没办法先当一个"搬运工”吧）。

 1 #include <iostream>
 2 #include <sstream>
 3 #include <cstdio>
 4 #include <climits>
 5 #include <cstring>
 6 #include <cstdlib>
 7 #include <string>
 8 #include <cmath>
 9 #include <stack>
10 #include <map>
11 #include <vector>
12 #include <queue>
13 #include <algorithm>
14 #define esp 1e-6
15 #define pb push_back
16 #define in  freopen("in.txt", "r", stdin);
17 #define out freopen("out.txt", "w", stdout);
18 #define print(a) printf("%d\n",(a));
19 #define bug puts("********))))))");
20 #define Rep(i, c) for(__typeof(c.end()) i = c.begin(); i != c.end(); i++)
21 #define inf 0x0f0f0f0f
22 using namespace std;
23 typedef long long  LL;
24 typedef vector<int> VI;
25 typedef pair<int, int> pii;
26 typedef vector<pii,int> VII;
27 typedef vector<int>:: iterator IT;
28 map<string, bool> my;
29 int ans;
30 void dfs(string cur)
31 {
32     if(my.find(cur) != my.end())
33         return;
34     int len = cur.size(), num = 0;
35     for(int i = 0; i < len; i++)
36         if(cur[i] == 'o')
37         num++;
38     ans = min(ans, num);
39     my[cur] = true;
40     for(int i = 0; i <= 9; i++)
41     {
42         if(cur.substr(i, 3) == "-oo")
43         {
44             string temp = cur;
45             temp.replace(i, 3, "o--");
46             dfs(temp);
47         }
48         if(cur.substr(i, 3) == "oo-")
49         {
50             string temp = cur;
51             temp.replace(i, 3, "--o");
52             dfs(temp);
53         }
54     }
55 }
56 int main(void)
57 {
58     int T;
59     string s;
60     for(int t = scanf("%d", &T); t <= T; t++)
61     {
62         cin>>s;
63         my.clear();
64         ans = 20;
65         dfs(s);
66         printf("%d\n", ans);
67     }
68     return 0;
69 }

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