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

这道题算是一道到隐式搜索题，将每一个状态转化为二进制，但是如果用将每个状态建成一个图，空间需要太大，不可行，所以运用，priority_queue优化的dijkstra算法，只需对现有的边进行转换即可

注意一点，优先队列优先最大，但是题目是优先最小，因为这样才能最优

这个题建图很有特色

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<queue>
#include<map>
#define MST(vis,x) memset(vis,x,sizeof(vis))
#define INF 0x3f3f3f3f
#define maxn  20
#define maxm 110
using namespace std;
int n,m;
char before[maxm][maxn+5],after[maxm][maxn+5];
int tim[maxm];
int d[(1<<maxn)+5];
int judge[(1<<maxn)+5];
struct node
{
    int bug;
    int dis;
    bool operator < (const node &x)const
    {
        return dis>x.dis;
    }
};
int dij()
{
    MST(d,INF);
    MST(judge,0);
    node temp,tt;
    temp.bug=(1<<n)-1;
    temp.dis=0;
    d[temp.bug]=0;
    priority_queue<node>q;
    q.push(temp);
    while(!q.empty())
    {
        temp=q.top();  q.pop();
        if(temp.bug==0)return temp.dis;
        if(judge[temp.bug])continue;
        judge[temp.bug]=1;
        for(int a=1;a<=m;a++)
        {
            int flag=0;
            for(int b=0;b<n;b++)
            {
                if(before[a][b]=='+'&&!(temp.bug&(1<<b))){flag=1;break;}
                if(before[a][b]=='-'&&(temp.bug&(1<<b))){flag=1;break;}
            }
            if(flag)continue;
            tt.bug=temp.bug;
            tt.dis=temp.dis+tim[a];
            for(int b=0;b<n;b++)
            {
                if(after[a][b]=='-')tt.bug&=~(1<<b);
                else if(after[a][b]=='+')tt.bug|=(1<<b);
            }
            int &D=d[tt.bug];
            if(tt.dis<D)
            {
                D=tt.dis;
                q.push(tt);
            }
        }
    }
    return -1;
}
int main()
{
    int ans,t=1;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(!n&&!m)break;
        for(int a=1;a<=m;a++)
            scanf("%d%s%s",&tim[a],before[a],after[a]);
        ans=dij();
        printf("Product %d\n",t++);
        if(ans<0)
            printf("Bugs cannot be fixed.\n\n");
        else printf("Fastest sequence takes %d seconds.\n\n",ans);
    }
    return 0;
}