題目鏈接

BZOJ3836

題解

顯然這是個\(NP\)完全問題，此題的解決全仗任意兩點間不存在節點數超過10的簡單路徑的性質
這意味著什麽呢？
\(dfs\)樹深度不超過\(10\)
\(10\)很小吶，可以狀壓了呢

我們發現一個點不但收祖先影響，而且受兒子影響，比較難處理
我們就先處理該點及其祖先，然後更新完兒子之後反過來用兒子更新根，就使得全局合法了
一個點顯然有三種狀態：
0.沒被覆蓋
1.被覆蓋但是沒有建站
2.建站

設\(f[d][s]\)表示節點\(u\)【深度為\(d\)】，其祖先【包括\(u\)】狀態為\(s\)的最優解
\(dfs\)進來的時候，我們用父親的答案更新\(u\)
\(dfs\)

復雜度\(O((n + m)3^{10})\)

#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\n",ans);
    return 0;
}

BZOJ3836 [Poi2014]Tourism 【樹形dp +狀壓dp】