Codeforces 161D Distance in Tree

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本文介绍了一种使用树形动态规划方法解决特定图论问题的算法。该算法用于计算一棵树中具有确切距离k的不同顶点对的数量。通过深度优先搜索进行两轮遍历，第一轮遍历计算每个节点到其子节点的距离分布，第二轮遍历计算满足条件的顶点对数量。

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A tree is a connected graph that doesn’t contain any cycles.

The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.

You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.

Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.

Next n - 1 lines describe the edges as “ai bi” (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.

Output
Print a single integer — the number of distinct pairs of the tree’s vertices which have a distance of exactly k between them.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Sample test(s)
input
5 2
1 2
2 3
3 4
2 5
output
4
input
5 3
1 2
2 3
3 4
4 5
output
2
Note
In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).

解题思路：简单树形DP。

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <utility>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long ll;
const int maxn = 50010;
ll dp1[maxn][510];
ll dp2[maxn];
int n, k;
ll  ans;

struct Edge {
    int v, next;
    Edge() { }
    Edge(int _v, int _next) : v(_v), next(_next) { }
}edges[2*maxn];
int head[maxn], edge_sum;

void init_graph() {
    edge_sum = 0;
    memset(head, -1, sizeof(head));
}

void add_edge(int u, int v) {
    edges[edge_sum].v = v;
    edges[edge_sum].next = head[u];
    head[u] = edge_sum++;

    edges[edge_sum].v = u;
    edges[edge_sum].next = head[v];
    head[v] = edge_sum++;
}

void dfs1(int u, int fa) {
    dp1[u][0] += 1;
    for(int i = head[u]; i != -1; i = edges[i].next) {
        int v = edges[i].v;
        if(v == fa) continue;
        dfs1(v, u);
        for(int j = 0; j < k; ++j) {
            dp1[u][j+1] += dp1[v][j];
        }
    }
    return ;
}

int par[maxn][510];

void dfs2(int u, int fa) {
    par[u][0] = u;
    dp2[u] = dp1[u][k];
    if(fa != -1) {
        for(int i = 1; i <= k; ++i) {
            par[u][i] = par[fa][i-1];
        }
        for(int i = 1; i <= k; ++i) {
            if(par[u][i] == -1) break;
            dp2[u] += dp1[par[u][i]][k-i];
            if(k-i-1 >= 0) {
                dp2[u] -= dp1[par[u][i-1]][k-i-1];
            }
        }
    }
    ans += dp2[u];
    for(int i = head[u]; i != -1; i = edges[i].next) {
        int v = edges[i].v;
        if(v == fa) continue;
        dfs2(v, u);
    }
    return ;
}


int main() {


    //freopen("aa.in", "r", stdin);

    int u, v;
    scanf("%d %d", &n, &k);
    init_graph();
    for(int i = 1; i < n; ++i) {
        scanf("%d %d", &u, &v);
        add_edge(u, v);
    }
    ans = 0;
    dfs1(1, -1);
    memset(par, -1, sizeof(par));
    dfs2(1, -1);
    printf("%I64d\n", ans/2);
    return 0;
}