ZOJ 1639 Hang Up the System （状态压缩）

Hang Up the System Time Limit: 2 Seconds       Memory Limit: 32768 KB You're going to design a multi-task operating system for an embedded system. Because the resources are limited on this mini computer, parallel running of some programs will probably give rise to the deadlock problem. According to experience, you have a list of programs that will hang up the system if all of them are running simultaneously. You are also given the priority of these programs. Now you're interested in the maximum value of the total priority of the programs that can run simultaneously in the system without hanging up.

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;
}