隐式的图搜索,存不下边,所以只有枚举转移就行了,因为bug的存在状态可以用二进制表示,转移的时候判断合法可以用位运算优化,
二进制pre[i][0]表示可以出现的bug,那么u&pre[i][0] == u就表示u是可以出现的bug集合的子集,
pre[i][1]表示必须出现的bug,那么u|pre[i][i] != u表示把必须出现的bug添加到u中,u中bug增加表面bug不全在u中,这是不合法的。
正权最短路就dijkstra,用spfa以前某题狂T有阴影。被输出格式坑得不要不要的,如果是if(kas) putchar('\n');就会WA...
#include<bits/stdc++.h>
using namespace std; const int maxm = ;
const int maxn = ; int pre[maxm][],nxt[maxm][];
int cost[maxm];
int n,m; int dist[<<maxn]; typedef pair<int,int> Node;
#define fi first
#define se second //bitset<20> temp;
#define bug(u)\
temp = u; cout<<#u<<'='<<temp<<endl;
#define cer(x)\
cout<<"dist="<<x<<endl; const int INF = 0x3f3f3f3f; void dijkstra()
{
priority_queue<Node,vector<Node>,greater<Node> > q;
memset(dist,0x3f,sizeof(int)*(<<n));
q.push(Node(,(<<n)-));
dist[(<<n)-] = ;
while(q.size()){
Node x = q.top(); q.pop();
if(x.se == ) { printf("Fastest sequence takes %d seconds.\n",dist[]); return; }
if(x.fi != dist[x.se]) continue;
int u = x.se;
for(int i = ; i < m; i++){
if( (pre[i][]&u) == u && (pre[i][]|u) == u){
int v = (u&nxt[i][])|nxt[i][];
if(dist[v] > dist[u]+cost[i]){
dist[v] = dist[u] + cost[i];
q.push(Node(dist[v],v));
}
}
}
}
puts("Bugs cannot be fixed.");
} int main()
{
//freopen("in.txt","r",stdin);
int kas = ;
char s1[maxn+],s2[maxn+];
while(scanf("%d%d",&n,&m),n){
for(int i = ; i < m ; i++){
scanf("%d%s%s",cost+i,s1,s2);
nxt[i][] = nxt[i][] = pre[i][] = pre[i][] = ;
for(int j = ; j < n; j++){
if(s1[j] == '+') pre[i][] |= <<j;
if(s1[j] != '-') pre[i][] |= <<j;
if(s2[j] == '+') nxt[i][] |= <<j;
if(s2[j] != '-') nxt[i][] |= <<j;
}
}
printf("Product %d\n",++kas);
dijkstra();
putchar('\n');
}
return ;
}