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  • HDU 4812 D Tree

    HDU 4812 

    思路:

    点分治

    先预处理好1e6 + 3以内到逆元

    然后用map 映射以分治点为起点的链的值a 成他的下标 u 

    然后暴力跑出以分治点儿子为起点的链的值b,然后在map里查找inv[b]*k

    代码:

    #include<bits/stdc++.h>
    using namespace std;
    #define fi first
    #define se second
    #define pi acos(-1.0)
    #define LL long long
    //#define mp make_pair
    #define pb push_back
    #define ls rt<<1, l, m
    #define rs rt<<1|1, m+1, r
    #define ULL unsigned LL
    #define pll pair<LL, LL>
    #define pii pair<int, int>
    #define mem(a, b) memset(a, b, sizeof(a))
    #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
    //head
    
    const int MOD = 1e6 + 3;
    const int INF = 0x7f7f7f7f;
    const int N = 1e5 + 5;
    int inv[MOD + 5], mp[MOD + 5], head[N], mxsz[N], sz[N], v[N], cnt = 0, rt = 0, n, k, ans1, ans2;
    int deep[N], dis[N], id[N], top = 0;
    bool vis[N];
    struct edge {
        int to, nxt;
    }edge[N*2];
    void add_edge(int u, int v) {
        edge[cnt].to = v;
        edge[cnt].nxt = head[u];
        head[u] = cnt++;
    }
    void init() {
        inv[1] = 1;
        for (int i = 2; i < MOD; i++) inv[i] = (MOD - MOD/i) * 1LL * inv[MOD%i] % MOD;
    }
    void update(int x, int y) {
        int t = (1LL * inv[x] * k) % MOD;
        int now = mp[t];
        if(!now) return ;
        if(now > y) swap(now, y);
        if(now < ans1 || now == ans1 && y < ans2) ans1 = now, ans2 = y;
    }
    void get_rt(int o, int u) {
        sz[u] = 1, mxsz[u] = 0;
        for (int i = head[u]; ~i; i = edge[i].nxt) {
            if(edge[i].to != o && !vis[edge[i].to]) {
                get_rt(u, edge[i].to);
                sz[u] += sz[edge[i].to];
                mxsz[u] = max(mxsz[u], sz[edge[i].to]);
            }
        }
        mxsz[u] = max(mxsz[u], n - sz[u]);
        if(mxsz[u] < mxsz[rt]) rt = u;
    }
    void get_d(int o, int u) {
        deep[++top] = dis[u];
        id[top] = u;
        for (int i = head[u]; ~i; i = edge[i].nxt) {
            if(!vis[edge[i].to] && edge[i].to != o) {
                dis[edge[i].to] = (1LL * dis[u] * v[edge[i].to])%MOD;
                get_d(u, edge[i].to);
            }
        }
    }
    void solve(int u) {
        vis[u] = true;
        mp[v[u]] = u;
        for (int i = head[u]; ~i; i = edge[i].nxt) {
            if(!vis[edge[i].to]) {
                top = 0, dis[edge[i].to] = v[edge[i].to];
                get_d(u, edge[i].to);
                for (int j = 1; j <= top; j++) update(deep[j], id[j]);
                top = 0, dis[edge[i].to] = (1LL * v[u] * v[edge[i].to])%MOD;
                get_d(u, edge[i].to);
                for (int j = 1; j <= top; j++) {
                    int t = deep[j];
                    if(!mp[t] || id[j] < mp[t]) mp[t] = id[j];
                }
            }
        }
        mp[v[u]] = 0;
        for (int i = head[u]; ~i; i = edge[i].nxt) {
            if(!vis[edge[i].to]) {
                top = 0, dis[edge[i].to] = (1LL * v[u] * v[edge[i].to])%MOD;
                get_d(u, edge[i].to);
                for (int j = 1; j <= top; j++) mp[deep[j]] = 0;
            }
        }
        for (int i = head[u]; ~i; i = edge[i].nxt) {
            if(!vis[edge[i].to]) {
                mxsz[0] = n =  sz[edge[i].to];
                get_rt(rt = 0, edge[i].to);
                solve(rt);
            }
        }
    }
    int main() {
        init();
        int u, V;
        while(~scanf("%d%d", &n, &k)) {
            mem(head, -1);
            mem(vis, false);
            mem(mp, 0);
            cnt = 0;
            ans1 = ans2 = INF;
            for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
            for (int i = 1; i < n; i++) scanf("%d%d", &u, &V), add_edge(u, V), add_edge(V, u);
            mxsz[0] = n;
            get_rt(rt = 0, 1);
            solve(rt);
            if(ans1 == INF) printf("No solution
    ");
            else printf("%d %d
    ", ans1, ans2);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/widsom/p/9106507.html
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