Input

The input has several test cases, each starts with n (2 <= n <= 16), the number of programs. The following n lines contain the name of the program and its priority, with each program on a line. The name is a string of no more than 10 characters. The priority is an integer from 1 to 100. The next line contains a single integer m (0 <= m <= 1,024), which is the length of the list of the conflicting programs. The following m lines each contains several names of the programs which cannot run together.

The input is terminated with a case n = 0. This case should not be processed.

Output

Print the case number and the maximal total priority value you can get on a line. Adhere to the sample output format.

Sample Input

3
HARDDISK 20
FLOPPY 10
CDROM 15
1
CDROM HARDDISK
5
HARDDISK 20
FLOPPY 10
CDROM 15
SERIAL 25
MOUSE 20
3
CDROM HARDDISK FLOPPY
FLOPPY SERIAL
SERIAL MOUSE
0

Sample Output

System 1: 30
System 2: 60

题意：先给出n个任务和每个任务的价值，然后给出m个限制条件，每个限制条件包括一些任务，说明这些任务不能同时进行处理。求能同时处理的任务的最大价值是多少。

分析：因为n不大于16，而且限制条件也比较少，所以很容易想到状态压缩。

先把那些限制条件转化为一个10进制整数，它的二进制形式中的1表示这些任务不能同时处理；然后从0开始枚举状态，判断当前状态与限制条件是否冲突，如果不冲突，算出当前状态的价值总和，与最大值进行比较即可。

注意：任务的名称不一定全是字母，可能由其他字符组成，我就是因为这里WA了还找不到错误。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<cstdlib>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<algorithm>
using namespace std;
typedef long long LL;

map<string, int> mp;
vector<int> v[1030];
int a[30];
int sta[1030];
int pro[17] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536};

int main()
{
    int n, m, i, j, cas = 0;
    string str, tmp;
    int val;
    while(cin >> n && n)
    {
        mp.clear();
        for(i = 1; i <= n; i++)
        {
            cin >> str >> val;
            mp[str] = i;
            a[i] = val;
        }
        memset(v, 0, sizeof(v));
        memset(sta, 0, sizeof(sta));
        cin >> m;
        getchar();
        for(int k = 0; k < m; k++)
        {
            getline(cin, str);
            int len = str.length();
            tmp = "";
            for(i = 0; i < len; i++)
            {
                if(str[i] != ' ' && str[i] != '\0') //注意此处的判断条件
                    tmp += str[i];
                else
                {
                    v[k].push_back(mp[tmp]);
                    tmp = "";
                }
            }
            v[k].push_back(mp[tmp]);
            sort(v[k].begin(), v[k].end());
            for(int j = 0; j < v[k].size(); j++)
                sta[k] += pro[n-v[k][j]];
        }
        int maxn = 1;
        for(i = 1; i <= n; i++)
            maxn *= 2;
        int ans = 0;
        for(int i = 0; i < maxn; i++)
        {
            int flag = 1;
            for(j = 0; j < m; j++)
            {
                if((i & sta[j]) == sta[j])
                {
                    flag = 0;
                    break;
                }
            }
            if(flag)
            {
                int tmp_sum = 0;
                int tt = 0, temp = i;
                while(temp)
                {
                    if(temp % 2 == 1)
                        tmp_sum += a[n-tt];
                    tt++;
                    temp /= 2;
                }
                ans = max(ans, tmp_sum);
            }
        }
        printf("System %d: %d\n", ++cas, ans);
    }
    return 0;
}