状压DP UVA 10817 Headmaster's Headache

 1 /*
 2     题意：学校有在任的老师和应聘的老师，选择一些应聘老师，使得每门科目至少两个老师教，问最少花费多少
 3     状压DP：一看到数据那么小，肯定是状压了。这个状态不好想，dp[s1][s2]表示s1二进制表示下至少有1位老师的科目集合
 4             s2表示至少有2位老师的科目集合所花费的最小金额，状态转移方程(01)：dp[t1][t2]=min(dp[t1][t2],dp[j][k]+c[i]);
 5             j,k为当前两个集合，t1，t2为转移后的集合，另外求t1，t2用到了& |位运算 1&1 == 1 1 & 0 == 0 0 & 0 == 0
 6             最后，学习了不知道数字多少个时应该用字符串整行读入
 7 　　详细解释
 8 
 9 */
10 /************************************************
11 * Author        :Running_Time
12 * Created Time  :2015-8-10 14:44:30
13 * File Name     :UVA_10817.cpp
14  ************************************************/
15 
16 #include <cstdio>
17 #include <algorithm>
18 #include <iostream>
19 #include <sstream>
20 #include <cstring>
21 #include <cmath>
22 #include <string>
23 #include <vector>
24 #include <queue>
25 #include <deque>
26 #include <stack>
27 #include <list>
28 #include <map>
29 #include <set>
30 #include <bitset>
31 #include <cstdlib>
32 #include <ctime>
33 using namespace std;
34 
35 #define lson l, mid, rt << 1
36 #define rson mid + 1, r, rt << 1 | 1
37 typedef long long ll;
38 const int MAXN = 150;
39 const int INF = 0x3f3f3f3f;
40 const int MOD = 1e9 + 7;
41 int c[MAXN], p[MAXN], cnt[10];
42 int dp[(1<<8)+10][(1<<8)+10];
43 int s, m, n;
44 int mxs;
45 int sum, st1, st2;
46 
47 int work(void)  {
48     memset (dp, INF, sizeof (dp));
49     dp[st1][st2] = sum;
50     for (int i=m+1; i<=m+n; ++i)    {
51         for (int j=mxs; j>=0; --j) {
52             for (int k=mxs; k>=0; --k) {
53                 if (dp[j][k] == INF)    continue;
54                 int t1 = (p[i] | j);  int t2 = (p[i] & j) | k;
55                 dp[t1][t2] = min (dp[t1][t2], dp[j][k] + c[i]);
56             }
57         }
58     }
59     return dp[mxs][mxs];
60 }
61 
62 int main(void)    {     //UVA 10817 Headmaster‘s Headache
63     while (scanf ("%d%d%d", &s, &m, &n) == 3)   {
64         if (!s) break;
65 
66         memset (cnt, 0, sizeof (cnt));
67         memset (p, 0, sizeof (p));
68         sum = 0, st1 = st2 = 0; mxs = (1 << s) - 1;
69         string str;
70         for (int i=1; i<=m+n; ++i)    {
71             scanf ("%d", &c[i]);
72             getline (cin, str);
73             for (int j=0; str[j]; ++j)  {
74                 if (isdigit (str[j]))   {
75                     int x = str[j] - ‘0‘;
76                     p[i] |= 1 << (x - 1);
77                     if (i <= m) ++cnt[x-1];
78                 }
79             }
80             if (i <= m) {
81                 sum += c[i];    st1 |= p[i];
82             }
83         }
84         for (int i=0; i<s; ++i) {
85             if (cnt[i] > 1) st2 |= (1 << i);
86         }
87 
88         printf ("%d\n", work ());
89     }
90 
91     return 0;
92 }