描述不清。。。
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<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
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<int> 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 3
1 000 00-
1 00- 0-+
2 0-- -++
*/