BZOJ3836 [Poi2014]Tourism 【树形dp +状压dp】

题目链接

BZOJ3836

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\n",ans);
	return 0;
}