洛谷P3806 点分治

点分治

第一次写点分治。。感觉是一个神奇而又暴力的东西orz

点分治大概就是用来处理树上链的信息，把路径分成过点x和不过点x的两种，不过点x的路径可以变成过点x的子树中一点的路径，递归处理

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 50005;
int n, m, cnt, rt, ans, tot, sum, head[N], size[N], dis[N], rd[N], query[N], all[N];
bool vis[N], judge[N], res[N];
struct Edge{ int v, next, w; } edge[N<<2];

void addEdge(int a, int b, int c){
    edge[cnt].v = b, edge[cnt].w = c, edge[cnt].next = head[a], head[a] = cnt ++;
}

void dfs(int s, int fa){
    size[s] = 1;
    int mp = 0;
    for(int i = head[s]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(u == fa || vis[u]) continue;
        dfs(u, s);
        size[s] += size[u];
        mp = max(mp, size[u]);
    }
    mp = max(mp, sum - size[s]);
    if(mp < ans) ans = mp, rt = s;
}

void getDis(int s, int fa){
    rd[++tot] = dis[s];
    for(int i = head[s]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(u == fa || vis[u]) continue;
        dis[u] = dis[s] + edge[i].w;
        getDis(u, s);
    }
}

void calc(int s){
    int num = 0;
    judge[0] = true;
    for(int i = head[s]; i != -1; i = edge[i].next){
        tot = 0; int u = edge[i].v;
        if(vis[u]) continue;
        dis[u] = edge[i].w, rd[0] = 0;
        getDis(u, s);
        for(int j = 1; j <= tot; j ++){
            for(int k = 0; k < m; k ++){
                if(query[k] >= rd[j]) res[k] |= judge[query[k] - rd[j]];
            }
        }
        for(int j = 1; j <= tot; j ++){
            all[++num] = rd[j], judge[rd[j]] = true;
        }
    }
    for(int i = 1; i <= num; i ++) judge[all[i]] = false;
}

void solve(int s){
    vis[s] = true, calc(s);
    for(int i = head[s]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(vis[u]) continue;
        ans = INF, sum = size[u];
        dfs(u, 0), solve(rt);
    }
}

int main(){

    full(head, -1);
    n = read(), m = read();
    for(int i = 0; i < n - 1; i ++){
        int u = read(), v = read(), c = read();
        addEdge(u, v, c), addEdge(v, u, c);
    }
    for(int i = 0; i < m; i ++) query[i] = read();
    ans = INF, sum = n;
    dfs(1, 0), solve(rt);
    for(int i = 0; i < m; i ++) printf(res[i] ? "AYE\n" : "NAY\n");
    return 0;
}