思路：通过前后两种状态建立一条边，利用Dijsktra就可以做了。
注意利用二进制优化。
AC代码
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = (1 << 20) + 5, maxm = 100 + 5;
int n, m, d[maxn], vis[maxn];
char before[maxm][25], after[maxm][25];
int cost[maxm];
struct Node{
int bug, dist;
Node(){}
Node(int bug, int dis):bug(bug),dist(dis){}
bool operator < (const Node& p) const {
return dist > p.dist;
}
};
int Dijsk(int u) {
memset(d, inf, sizeof(d));
memset(vis, 0, sizeof(vis));
priority_queue<Node>Q;
Q.push(Node(u, 0));
d[u] = 0;
while(!Q.empty()) {
Node p = Q.top(); Q.pop();
int u = p.bug;
if(u == 0) return d[u];
if(vis[u]) continue;
vis[u] = 1;
for(int i = 0; i < m; ++i) {
bool ok = true;
for(int j = 0; j < n; ++j) {
if(before[i][j] == '+' && !(u & (1 << j))) {ok = false; break;}
if(before[i][j] == '-' && (u & (1 << j))) {ok = false; break;}
}
if(!ok) continue; //不能打补丁
Node v = Node(u, p.dist + cost[i]);
for(int j = 0; j < n; ++j) {
if(after[i][j] == '-') v.bug &= ~(1 << j);
if(after[i][j] == '+') v.bug |= (1 << j);
}
if(v.dist < d[v.bug] || d[v.bug] < 0) {
d[v.bug] = v.dist;
Q.push(v);
}
}
}
return -1;
}
int main() {
int kase = 0;
while(scanf("%d%d", &n, &m) == 2 && n && m) {
for(int i = 0; i < m; ++i) {
scanf("%d%s%s", &cost[i], before[i], after[i]);
}
int ans = Dijsk((1 << n) - 1);
printf("Product %d\n", ++kase);
if(ans == -1) printf("Bugs cannot be fixed.\n");
else printf("Fastest sequence takes %d seconds.\n", ans);
printf("\n");
}
return 0;
}如有不当之处欢迎指出！