UVA - 658 It's not a Bug, it's a Feature!

题意：某个软件有n个bug（n<=20）和m个补丁（m<=100），每个补丁可以修复某些bug，同时又会带来某些bug，而且每个补丁要能起效都必须满足某些条件：在安装此补丁前，必须有某些指定bug存在以及某些指定的bug不存在。

每个补丁都有一个安装时间，问最短需要多少时间才可以消除所有bug?

参考了大神的二进制表示的方法

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 110;
const int MAXD = 2000000;
const int INF = 0x3f3f3f3f;

int s[2][MAXN],t[2][MAXN],w[MAXN],N,M;
char a[MAXN],b[MAXN];
int q[MAXD],inq[MAXD],d[MAXD];

int init(){
    scanf("%d%d",&N,&M);
    if (!M && !N)
        return 0;
    memset(s,0,sizeof(s));
    memset(t,0,sizeof(t));
    for (int i = 0; i < M; i++){
        scanf("%d%s%s",&w[i],a,b);
        for (int j = 0; j < N; j++){
            if (a[j] == '+')
                s[1][i] |= (1<<j);
            if (a[j] == '-')
                s[0][i] |= (1<<j);
            if (b[j] == '+')
                t[1][i] |= (1<<j);
            if (b[j] == '-')
                t[0][i] |= (1<<j);
        }
    }
    return 1;
}

void SPFA(){
    int Max = 1<<N;
    for (int i = 0; i < Max; i++){
        d[i] = INF;
        inq[i] = 0;
    }
    int front=0,rear=0;
    d[Max-1] = 0;
    q[rear++] = Max - 1;
    while (front != rear){
        int u = q[front++];
        if (front > Max)
            front = 0;
        inq[u] = 0;
        for (int i = 0; i < M; i++)
            if ((u & s[1][i]) == s[1][i] && ((~u) & s[0][i]) == s[0][i]){
                int v = u;
                v |= t[1][i];
                v &= (~t[0][i]);
                if (d[u]+w[i] < d[v]){
                    d[v] = d[u] + w[i];
                    if (!inq[v]){
                        q[rear++] = v;
                        if (rear > Max)
                            rear = 0;
                        inq[v] = 1;
                    }
                }
            }
    }
    if (d[0] == INF)
       puts("Bugs cannot be fixed.");
    else printf("Fastest sequence takes %d seconds.\n",d[0]);
}

int main(){
    int t = 0;
    while (init()){
        printf("Product %d\n",++t);
        SPFA();
        printf("\n");
    }
    return 0;
}