Codeforces 711C. Bear and Tree Jumps【树形dp好题】

题目：http://codeforces.com/contest/791/problem/D

题意：

你可以从树上的节点一次最多走k条边。
(称为跳一次);
树为无权树;
然后问你任意两点之间的条的次数的和为多少;

分析：

不会做，看题解：http://codeforces.com/blog/entry/51068

代码：

#include <bits/stdc++.h>using namespace std;const int nax = 2e5 + 5;vector<int> edges[nax];int count_subtree[nax][5];int total_subtree[nax];long long answer;int n, k;int subtract(int a, int b) {    return ((a - b) % k + k) % k;}void dfs(int a, int par, int depth) {    count_subtree[a][depth % k] = total_subtree[a] = 1;    for(int b : edges[a])        if(b != par) {            dfs(b, a, depth + 1);            for(int i = 0; i < k; ++i)                for(int j = 0; j < k; ++j) {                    // compute distance modulo k                    int remainder = subtract(i + j, 2 * depth);                    // compute x such that (distance + x) is divisible by k                    int needs = subtract(k, remainder);                    answer += (long long) needs                            * count_subtree[a][i] * count_subtree[b][j];                }            for(int i = 0; i < k; ++i)                count_subtree[a][i] += count_subtree[b][i];            total_subtree[a] += total_subtree[b];        }    // in how many pairs we will count the edge from 'a' to its parent?    answer += (long long) total_subtree[a] * (n - total_subtree[a]);}int main() {    scanf("%d%d", &n, &k);    for(int i = 0; i < n - 1; ++i) {        int a, b;        scanf("%d%d", &a, &b);        edges[a].push_back(b);        edges[b].push_back(a);    }    dfs(1, -1, 0);    assert(answer % k == 0);    printf("%lld\n", answer / k);}