题目链接
题解
显然这是个(NP)完全问题，此题的解决全仗任意两点间不存在节点数超过10的简单路径的性质
这意味着什么呢？
(dfs)树深度不超过(10)
(10)很小呐，可以状压了呢
我们发现一个点不但收祖先影响，而且受儿子影响，比较难处理
我们就先处理该点及其祖先，然后更新完儿子之后反过来用儿子更新根，就使得全局合法了
一个点显然有三种状态：
0.没被覆盖
1.被覆盖但是没有建站
2.建站
设(f[d][s])表示节点(u)【深度为(d)】，其祖先【包括(u)】状态为(s)的最优解
(dfs)进来的时候，我们用父亲的答案更新(u)
(dfs)结束的时候，我们用儿子的答案替代(u)的答案，保证全局合法
复杂度(O((n + m)3^{10}))
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 20005,maxm = 50005,M = 59050,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int h[maxn],ne = 1;
int n,m,C[maxn],dep[maxn],vis[maxn],bin[100];
int f[11][M],st[maxn],top; //0 not yet 1 ok but empty 2 ok and full
struct EDGE{int to,nxt;}ed[maxm];
inline void build(int u,int v){
ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;
ed[++ne] = (EDGE){u,h[v]}; h[v] = ne;
}
inline int min(int a,int b){return a < b ? a : b;}
void dfs(int u){
vis[u] = true;
int maxv = bin[dep[u]] - 1,d = dep[u]; top = 0;
for (int i = 0; i < bin[dep[u] + 1]; i++) f[d][i] = INF;
Redge(u) if (vis[to = ed[k].to]) st[++top] = dep[to];
if (!d) f[0][0] = 0,f[0][1] = INF,f[0][2] = C[u];
for (int s = 0; s <= maxv; s++){
int t = 0,v,p,e = s + 2 * bin[d];
REP(i,top){
v = st[i]; p = s / bin[v] % 3;
if (p == 2) t = 1;
else if (!p) e += bin[v];
}
f[d][s + t * bin[d]] = min(f[d][s + t * bin[d]],f[d - 1][s]);
f[d][e] = min(f[d][e],f[d - 1][s] + C[u]);
}
Redge(u) if (!vis[to = ed[k].to]){
dep[to] = dep[u] + 1;
dfs(to);
for (int s = 0; s < bin[d + 1]; s++)
f[d][s] = min(f[d + 1][s + bin[d + 1]],f[d + 1][s + 2 * bin[d + 1]]);
}
}
int main(){
bin[0] = 1; for (int i = 1; i <= 13; i++) bin[i] = bin[i - 1] * 3;
n = read(); m = read();
REP(i,n) C[i] = read();
while (m--) build(read(),read());
int ans = 0;
for (int i = 1; i <= n; i++)
if (!vis[i]) dfs(i),ans += min(f[0][1],f[0][2]);
printf("%d
",ans);
return 0;
}