Pi of each job seeker, for each job seeker, either or not, is the 01 backpack problem.
For s1 and s2, you can follow the courses of the currently enumerated job seekers to publish the next status:
NextS1 = p [I] | s1,
NextS2 = (p [I] & s1) | s2
F [nextS1] [nextS2] = min (f [nextS1] [nextS2], f [s1] [s2] + p [I])
/**========================================== * This is a solution for ACM/ICPC problem * * @problem: UVA 10817 - Headmaster's Headache * @type: dp * @author: shuangde * @blog: blog.csdn.net/shuangde800 * @email: zengshuangde@gmail.com *===========================================*/#include<iostream>#include<cstdio>#include<algorithm>#include<vector>#include<queue>#include<cmath>#include<cstring>using namespace std;typedef long long int64;const int INF = 0x3f3f3f3f;const double PI = acos(-1.0);const int MAXN = 150;int courseNum, m, n, sum; int maxState;int f[1<<8][1<<8];int p[MAXN], c[MAXN], cnt[10];int dp(int st1, int st2){ memset(f, 0x3f, sizeof(f)); f[st1][st2] = sum; for(int i=m+1; i<=n+m; ++i){ for(int s1=maxState; s1>=0; --s1){ for(int s2=maxState; s2>=0; --s2){ if(f[s1][s2] >= INF) continue; int st1 = (p[i]|s1); int st2 = (p[i]&s1) | s2; f[st1][st2] = min(f[st1][st2], f[s1][s2]+c[i]); } } } return f[maxState][maxState];}int main(){ char str[1000]; while(~scanf("%d%d%d", &courseNum, &m, &n) && courseNum){ maxState = (1<<courseNum) - 1; sum = 0; int st1=0, st2=0; memset(cnt, 0, sizeof(cnt)); for(int i=1; i<=m+n; ++i){ scanf("%d", &c[i]); gets(str); p[i] = 0; for(int j=0; j<strlen(str); ++j){ if(isdigit(str[j])){ int num = str[j] - '0'; p[i] |= 1<<(num-1); if(i <= m) ++cnt[num-1]; } } if(i <= m){ sum += c[i]; st1 |= p[i]; } } for(int i=0; i<courseNum; ++i) if(cnt[i] > 1) st2 |= (1<<i); printf("%d\n", dp(st1, st2)); } return 0;}