WOJ 1605 Distance on Tree（莫队算法）

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题意：给出有n个结点的一棵树，并给出参数k，除了0结点外每个节点i与I/k连有一条权值为i的边，现在有q组询问，每组询问查询[I,j]中任意两个节点之间的距离之和。

思路：假设当前查询区间为[L,R],那么每个结点u往上连的那条边对答案的贡献度为d[u]*(R-L+1)*u,其中d[u]表示u有多少子孙结点在[L,R]中，

然后对于这n-1条边的贡献度求和即可。

然后就可以用莫队算法来做了，首先因为每次查询区间的大小是变的，所以用一个变量sum_d_mul_i记录当前sigma(d[u]*I),然后莫队的时候动态维护sum_d_mul_i和d[I]即可。

具体见代码：

#include<bits/stdc++.h>#define eps 1e-6#define LL long long#define pii pair<int, int>#define pb push_back#define mp make_pair//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;const int MAXN = 10010;//const int INF = 0x3f3f3f3f;int n, k, q, unit;int d[MAXN];LL ans[MAXN], ret, sum_d_mul_i, len;struct Query{int l, r;int id;} query[MAXN];bool cmp(const Query& A, const Query& B) {if (A.l/unit != B.l/unit) return A.l/unit < B.l/unit;return A.r < B.r;}void inc(int u) {while (u) {ret -= d[u]*(len-d[u])*u;d[u]++;ret += d[u]*(len-d[u])*u;sum_d_mul_i += u;u /= k;}}void dec(int u) {while (u) {ret -= d[u]*(len-d[u])*u;d[u]--;ret += d[u]*(len-d[u])*u;sum_d_mul_i -= u;u /= k;}}void work() {ret = sum_d_mul_i = 0;len = 1;int L = 0, R = 0;for (int i = 1; i <= q; i++) {while (R < query[i].r) { len++;ret += sum_d_mul_i;R++;inc(R); }while (R > query[i].r) {len--;ret -= sum_d_mul_i;dec(R); R--;}while (L < query[i].l) {len--;ret -= sum_d_mul_i;dec(L); L++;}while (L > query[i].l) {len++;ret += sum_d_mul_i;L--;inc(L);}ans[query[i].id] = ret;}}int main(){    //freopen("input.txt", "r", stdin);while (scanf("%d%d%d", &n, &k, &q) == 3) {memset(d, 0, sizeof(d));for (int i = 1; i <= q; i++) {scanf("%d%d", &query[i].l, &query[i].r);query[i].id = i;}query[0].id = 0, query[0].l = 0, query[0].r = 0;unit = (int)sqrt(n);sort(query, query+q+1, cmp);work();for (int i = 1; i <= q; i++)printf("%I64d\n", ans[i]);}    return 0;}

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