题意

描述不清。。。

Sol

网络流24题里面怎么会有状压dp？？

真是狗血，不过还是简单吧。

直接用$f[sta]$表示当前状态为$sta$时的最小花费

转移的时候枚举一下哪一个补丁可以搞这个状态

但是这玩意儿有后效性，可以用SPFA消去

 

#include
     
      #include
      
       #include
       
        #include
        
         using namespace std;const int MAXN = 101, INF = 1e9 + 10;inline int read() {    char c = getchar(); int x = 0, f = 1;    while(c < '0' || c > '9') {
   if(c == '-') f = 1; c = getchar();}    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();    return x * f;}int N, M;int tim[MAXN], dis[2097153], B1[MAXN], B2[MAXN], F1[MAXN], F2[MAXN], vis[2097153];int SPFA() {    queue
         
           q;     q.push((1 << N) - 1); dis[(1 << N) - 1] = 0;    while(!q.empty()) {        int sta = q.front(); q.pop(); vis[sta] = 0;        for(int i = 1; i <= M; i++) {            int now = sta;            if(((sta & B1[i]) == B1[i]) && ((sta & B2[i]) == 0)) {                now = now & (~F1[i]);                now = now | F2[i];                if(dis[now] > dis[sta] + tim[i]) {                    dis[now] = dis[sta] + tim[i];                    if(!vis[now]) q.push(now);                }            }        }    }    return dis[0] < INF ? dis[0] : 0;}int main() {    N = read(); M = read();    memset(dis, 0x3f, sizeof(dis));    for(int i = 1; i <= M; i++) {        char s1[21], s2[21];        scanf("%d %s %s", &tim[i], s1, s2);        for(int j = 0; j < N; j++)            if(s1[j] == '+') B1[i] |= 1 << j;            else if(s1[j] == '-') B2[i] |= 1 << j;        for(int j = 0; j < N; j++)            if(s2[j] == '-') F1[i] |= 1 << j;            else if(s2[j] == '+') F2[i] |= 1 << j;    }    printf("%d", SPFA());    return 0;}/*3 31 000 00-1 00- 0-+2 0-- -++*/