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  • LOJ 2491 求和 (LCA + 前缀和)

    Loj.2491

    题意:
    给一棵有根树,对Q次询问,每次输入x,y,k。输出树上x到y的路径上点的深度的k次方和。

    思路:
    树上两点间路径的权值和很容易想到LCA, 然后发现可以预处理深度的k次方的前缀和。对每个x和lca之间点的深度肯定是连续和,其深度k次方和(不算lca点)是sum[d[x]][k] - sum[d[lca]][k]。

    代码:

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    #include<map>
    #include<queue>
    #include<vector>
    #include<string>
    #include<bitset>
    #include<fstream>
    using namespace std;
    #define rep(i, a, n) for(int i = a; i <= n; ++ i);
    #define per(i, a, n) for(int i = n; i >= a; -- i);
    typedef long long ll;
    typedef pair<int,int> PII;
    const int N = 1e6 + 105;
    const int mod = 998244353;
    const double Pi = acos(- 1.0);
    const int INF = 0x3f3f3f3f; 
    const int G = 3, Gi = 332748118;
    ll qpow(ll a, ll b) { ll res = 1; while(b){ if(b & 1) res = (res * a) % mod; a = (a * a) % mod; b >>= 1;} return res; }
    ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
    // bool cmp(int a, int b){return a > b;}
    //
    
    int n;
    int head[N], cnt = 0;
    int to[N << 1], nxt[N << 1];
    ll sum[N][60];
    
    void add(int u, int v){
        to[cnt] = v, nxt[cnt] = head[u], head[u] = cnt ++;
        to[cnt] = u, nxt[cnt] = head[v], head[v] = cnt ++;
    }
    
    
    //lca
    int t;
    int d[N], dist[N], f[N][20];
    
    queue<int> q;
    
    void bfs(){
        q.push(1);
        d[1] = 0;
        while(q.size()){
            int u = q.front(); q.pop();
            for(int i = head[u]; i != -1; i = nxt[i]){
                int v = to[i];
                if(d[v] != -1) continue;
                d[v] = d[u] + 1;
                f[v][0] = u;
                for(int j = 1; j <= t; ++ j)
                    f[v][j] = f[f[v][j - 1]][j - 1];
                q.push(v);
            }
        }
    }
    
    int Lca(int x,int y)
    {
        //调整到同样高度
        if(d[x] > d[y]) swap(x, y);
        for(int i = t; i >= 0; -- i)
            if(d[f[y][i]] >= d[x]) y = f[y][i];
        //特殊情况
        if(x == y) return x;
        //一般情况
        for(int i = t; i >= 0; -- i)
            if(f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
        return f[x][0];
    }
    
    
    int main()
    {
        scanf("%d",&n);
        cnt = 0;
        memset(d, -1, sizeof(d));
        for(int i = 0; i <= n; ++ i) head[i] = -1;
        for(int i = 1; i <= n; ++ i){
            ll tp = 1;
            for(int j = 1; j <= 50; ++ j){
                tp = tp * 1ll * i % mod;
                sum[i][j] = (sum[i - 1][j] + tp) % mod;
            }
        }
        for(int i = 1; i < n; ++ i){
            int x, y; scanf("%d%d",&x,&y);
            add(x, y);
        }
        t=(int)(log(n)/log(2))+1;
        bfs();
        int Q; scanf("%d",&Q);
        while(Q --){
            int x, y; ll k; scanf("%d%d%lld",&x,&y,&k);
            int lca = Lca(x, y);
            ll res = ((sum[d[x]][k] + sum[d[y]][k] - sum[d[lca]][k] * 2ll % mod + mod) % mod + qpow(d[lca], k)) % mod; 
            printf("%lld
    ",res);
        }
        return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/A-sc/p/13758228.html
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