BZOJ3836 [Poi2014]Tourism 【树形dp +状压dp】
2018-06-29 12:08
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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\n",ans); return 0; }
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