SPOJ 1825 Free tour II 解题报告（树分治）

1825. Free tour II

Problem code: FTOUR2

After the success of 2nd anniversary (take a look at problem FTOUR for more details), this 3rd year, Travel Agent SPOJ goes on with another discount tour.

The tour will be held on ICPC island, a miraculous one on the Pacific Ocean. We list N places (indexed from 1 to N) where the visitors can have a trip. Each road connecting them has an interest value, and this value can be negative(if there is nothing interesting to view there). Simply, these N places along with the roads connecting them form atree structure. We will choose two places as the departure and destination of the tour.

Since September is the festival season of local inhabitants, some places are extremely crowded (we call themcrowded places). Therefore, the organizer of the excursion hopes the tour will visit at most K crowded places (too tiring to visit many of them) and of course, the total number of interesting value should be maximum.

Briefly, you are given a map of N places, an integer K, and M id numbers of crowded place. Please help us to find the optimal tour. Note that we can visit each place only once (or our customers easily feel bored), also the departure and destination places don't need to be different.

Input

There is exactly one case. First one line, containing 3 integers N K M, with 1 <= N <= 200000, 0 <= K <= M, 0 <=M <= N.

Next M lines, each line includes an id number of a crowded place.

The last (N - 1) lines describe (N - 1) two-way roads connected N places, form a b i, with a, b is the id of 2 places, and i is its interest value (-10000 <= i <= 10000).

Output

Only one number, the maximum total interest value we can obtain.

Example

Input:8 2 33571 3 12 3 103 4 -24 5 -15 7 65 6 54 8 3Output:12

    解题报告：蛮艰难的一道题。漆子超论文第二题，思路直接看他的就好了，懂了以后就可以敲代码了。注意很多细节，问题应该不大。代码如下：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <queue>#include <vector>#include <map>#include <set>#include <string>#include <iomanip>#include <cassert>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000")#define ff(i, n) for(int i=0;i<(n);i++)#define fff(i, n, m) for(int i=(n);i<=(m);i++)#define dff(i, n, m) for(int i=(n);i>=(m);i--)#define travel(e, u) for(int e = u, v = vv[u]; e; e = nxt[e], v = vv[e])#define Bit(n) (1LL<<(n))#define And(a, b) ((a) & (b))#define Xor(a, b) ((a) ^ (b))#define Or(a, b) ((a) | (b))typedef long long LL;typedef unsigned long long ULL;void work();int main(){#ifdef ACM    freopen("in.txt", "r", stdin);#endif // ACM    work();}void scanf(int & x, char c = 0){    while((c = getchar()) < '0' || c > '9');    x = c - '0';    while((c = getchar()) >= '0' && c <= '9') x = x * 10 + (c - '0');}#define rep(i,a,b) for(int i = (a);i<(b);++i)#define rrep(i,a,b) for(int i = (a);i>=(b);--i)#define clr(a,x) memset(a,x,sizeof(a))/*****************************************华丽分割线**********************************************/const int maxv = 222222;/// Globalint n, m, k;int ans;/// Edgeint edge[maxv], ecnt;int nxt[maxv * 2], vv[maxv * 2], ww[maxv * 2];/// Treebool blk[maxv], vis[maxv];int siz[maxv], mon[maxv], dad[maxv];int root, minN, maxD;/// Tree Nodeint all[maxv], allK;int now[maxv], nowK;struct Node{    int v, w, d;    Node(int v = 0, int w = 0, int d = 0) : v(v), w(w), d(d) {}    bool operator<(const Node & cmp) const    {        return d < cmp.d;    }};void init(){    /// Edge initial    clr(edge, 0);    ecnt = 2;    /// Ans Initial    ans = 0;    /// Tree Initial    clr(blk, 0);    clr(vis, 0);}void addEdge(int u, int v, int w, int first[]){    nxt[ecnt] = first[u], vv[ecnt] = v, ww[ecnt] = w, first[u] = ecnt++;}void input(){    ff(i, m)    {        int u;        scanf("%d", &u);        blk[u] = true;    }    ff(i, n-1)    {        int u, v, w;        scanf("%d%d%d", &u, &v, &w);        addEdge(u, v, w, edge);        addEdge(v, u, w, edge);    }}void dfsSize(int u){    siz[u] = 1;    mon[u] = 0;    travel(e, edge[u]) if(!vis[v] && v != dad[u])        dad[v] = u, dfsSize(v), siz[u] += siz[v], mon[u] = max(mon[u], siz[v]);}int r;void dfsRoot(int u){    mon[u] = max(mon[u], siz[r] - siz[u]);    if(mon[u] < minN) minN = mon[u], root = u;    travel(e, edge[u]) if(!vis[v] && v != dad[u])        dfsRoot(v);}int bb;void dfsDeep(int u){    if(blk[u]) bb++;    maxD = max(maxD, bb);    travel(e, edge[u]) if(!vis[v] && v != dad[u])        dad[v] = u, dfsDeep(v);    if(blk[u]) bb--;}int wv;void dfsValue(int u){    if(blk[u]) bb++;    now[bb] = max(now[bb], wv);    travel(e, edge[u]) if(!vis[v] && v != dad[u])        wv += ww[e], dfsValue(v), wv -= ww[e];    if(blk[u]) bb--;}void dfs(int u){    /// Get root    dad[u] = 0, dfsSize(u);    r = u, minN = n, dfsRoot(u);    /// Get son and sort    u = root;    vector<Node> son;    travel(e, edge[u]) if(!vis[v])        bb = 0, dad[v] = u, maxD = 0, dfsDeep(v), son.push_back(Node(v, ww[e], maxD));    sort(son.begin(), son.end());    if(son.size() == 0) return;    allK = -1;    int kk = k;    if(blk[u]) kk--;    if(kk >= 0)        ff(i, son.size())    {        nowK = son[i].d;        fff(j, 0, nowK) now[j] = Bit(31);        bb = 0;        wv = son[i].w;        dfsValue(son[i].v);        fff(j, 1, nowK) now[j] = max(now[j], now[j-1]);        ans = max(ans, now[min(kk, nowK)]);        if(allK == -1)        {        }        else if(allK < kk)        {            if(allK + nowK <= kk)                ans = max(ans, now[nowK] + all[allK]);            else                fff(j, max(0, kk-nowK), allK)                    ans = max(ans, all[j] + now[kk-j]);        }        else        {            dff(j, kk, 0)                ans = max(ans, all[j] + now[kk-j]);        }        /// Merge        fff(j, 0, allK) all[j] = max(all[j], now[j]);        if(allK >= 0) all[0] = max(all[0], now[0]);        fff(j, max(allK, 0) + 1, nowK) all[j] = max(now[j], all[j-1]);        allK = nowK;    }    vis[u] = true;    travel(e, edge[u]) if(!vis[v])        dfs(v);}void work(){    while(scanf("%d%d%d", &n, &k, &m) == 3)    {        init();        input();        dfs(1);        printf("%d\n", ans);    }}