It's not a Bug, It's a Feature! (poj 1482 最短路SPFA+隐式图+位运算)

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It's not a Bug, It's a Feature!

Time Limit: 5000MS		Memory Limit: 30000K
Total Submissions: 1353		Accepted: 516

Description

It is a curious fact that consumers buying a new software product generally do not expect the software to be bug-free. Can you imagine buying a car whose steering wheel only turns to the right?

Or a CD-player that plays only CDs with country music on them?
Probably not. But for software systems it seems to be acceptable if they do not perform as they should do. In fact, many software companies have adopted the habit of sending out patches to fix bugs every few weeks after a new product is released (and even
charging money for the patches). 

Tinyware Inc. is one of those companies. After releasing a new word processing software this summer, they have been producing patches ever since. Only this weekend they have realized a big problem with the patches they released. While all patches fix some bugs,
they often rely on other bugs to be present to be installed. This happens because to fix one bug, the patches exploit the special behavior of the program due to another bug. 



More formally, the situation looks like this. Tinyware has found a total of n bugs B = {b1, b2, ..., bn} in their software. And they have released m patches p1, p2, ..., pm. To apply patch pi to the software, the bugs Bi⁺ in B have to be present
in the software, and the bugs Bi^- in B must be absent (of course Bi⁺ ∩ Bi^- = Φ). The patch then fixes the bugs Fi^- in B (if they have been present) and introduces the new bugs Fi⁺ in B (where, again, Fi⁺∩
Fi^- = Φ). 



Tinyware's problem is a simple one. Given the original version of their software, which contains all the bugs in B, it is possible to apply a sequence of patches to the software which results in a bug- free version of the software? And if so, assuming that
every patch takes a certain time to apply, how long does the fastest sequence take?

Input

The input contains several product descriptions. Each description starts with a line containing two integers n and m, the number of bugs and patches, respectively. These values satisfy 1 <= n <= 20 and 1 <= m <= 100. This is followed by m lines describing the
m patches in order. Each line contains an integer, the time in seconds it takes to apply the patch, and two strings of n characters each. 



The first of these strings describes the bugs that have to be present or absent before the patch can be applied. The i-th position of that string is a ``+'' if bug bi has to be present, a ``-'' if bug bi has to be absent, and a `` 0'' if it doesn't matter whether
the bug is present or not. 



The second string describes which bugs are fixed and introduced by the patch. The i-th position of that string is a ``+'' if bug bi is introduced by the patch, a ``-'' if bug bi is removed by the patch (if it was present), and a ``0'' if bug bi is not affected
by the patch (if it was present before, it still is, if it wasn't, is still isn't). 



The input is terminated by a description starting with n = m = 0. This test case should not be processed. 

Output

For each product description first output the number of the product. Then output whether there is a sequence of patches that removes all bugs from a product that has all n bugs. Note that in such a sequence a patch may be used multiple times. If there is such
a sequence, output the time taken by the fastest sequence in the format shown in the sample output. If there is no such sequence, output ``Bugs cannot be fixed.''. 



Print a blank line after each test case.

Sample Input

3 3
1 000 00-
1 00- 0-+
2 0-- -++
4 1
7 0-0+ ----
0 0

Sample Output

Product 1
Fastest sequence takes 8 seconds.

Product 2
Bugs cannot be fixed.

Source

Southwestern European Regional Contest 1998

这个题拿到手并不会做，没有好的思路，然后就看了网上的题解。第一次碰到不建图也能SPFA的，又学习了。另外这一题的位运算处理也非常巧妙，这是我不熟悉的，先放在这里，以后多来看几遍。

參考这两篇博客，写的非常好：http://www.cnblogs.com/scau20110726/archive/2012/12/16/2820739.html

http://www.cnblogs.com/staginner/archive/2011/10/25/2223489.html

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 25;
const int MAXN = 111;
const int MAXM = (1<<20)+100;
const int N = 1005;

char s1[maxn],s2[maxn];
int s[2][MAXN],t[2][MAXN],cost[MAXN];
int dist[MAXM];
bool inq[MAXM];
int n,m;

void SPFA()
{
    int i,j;
    mem(inq,false);
    mem(dist,INF);
    queue<int>Q;
    int start=(1<<n)-1;
    Q.push(start);
    inq[start]=true;
    dist[start]=0;
    while (!Q.empty())
    {
        int u=Q.front(); Q.pop();
        inq[u]=false;
        for (i=0;i<m;i++)
        {
            if ((u|s[1][i])==u&&(u&s[0][i])==u)
            {
                int v=u;
                v|=t[1][i];
                v&=t[0][i];
                if (dist[v]>dist[u]+cost[i])
                {
                    dist[v]=dist[u]+cost[i];
                    if (!inq[v])
                    {
                        inq[v]=true;
                        Q.push(v);
                    }
                }
            }
        }
    }
//    for(i=0;i<=start;i++)
//        pf("%d ",dist[i]);
//    pf("\n");
    if (dist[0]==INF)
        pf("Bugs cannot be fixed.\n");
    else
        pf("Fastest sequence takes %d seconds.\n",dist[0]);
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
    int i,j,cas=0;
    while (sff(n,m))
    {
        if (n==0&&m==0) break;
        mem(s,0);
        mem(t,0);
        for (i=0;i<m;i++)
        {
            scanf("%d%s%s",&cost[i],s1,s2);
            for (j=0;j<n;j++)
            {
                if (s1[j]=='+')
                    s[1][i]+=(1<<j);
                if (s1[j]!='-')
                    s[0][i]+=(1<<j);
                if (s2[j]=='+')
                    t[1][i]+=(1<<j);
                if (s2[j]!='-')
                    t[0][i]+=(1<<j);
            }
        }
        pf("Product %d\n",++cas);
        SPFA();
        pf("\n");
    }
    return 0;
}