CodeForces 161D Distance in Tree（树形DP）

最新推荐文章于 2024-12-07 12:33:10 发布

Tongqi_Liu

最新推荐文章于 2024-12-07 12:33:10 发布

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本文链接：https://blog.youkuaiyun.com/fipped/article/details/47685597

本文详细解析了一道关于在树形结构中计算长度为K的路径数量的编程题，通过动态规划方法逐步深入理解解题思路。

题目链接：http://codeforces.com/problemset/problem/161/D

题意：给一棵树，统计树中长度为K的路径的数目

思路：dp[u][x]表示通过u点且长为x的路径数目，为防止计数重复需要将计数提前到更新答案前

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <climits>
#include <functional>
#include <deque>
#include <ctime>
#include <string>

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#pragma comment(linker, "/STACK:102400000,102400000")

using namespace std;

const int maxn = 50010;

int cnt, head[maxn];
int dp[maxn][510], ans;

struct Edge
{
    int to, next;
} edge[maxn << 1];

void init()
{
    cnt = 0;
    memset(head, -1, sizeof(head));
}

void addedge(int u, int v)
{
    edge[cnt].to = v;
    edge[cnt].next = head[u];
    head[u] = cnt++;
}

void dfs(int u, int fa, int k)
{
    dp[u][0] = 1;
    for (int i = 1; i <= k; i++)
        dp[u][i] = 0;

    for (int i = head[u]; ~i; i = edge[i].next)
    {
        int v = edge[i].to;
        if (v == fa) continue;

        dfs(v, u, k);
        for (int j = 0; j < k; j++)
            ans += dp[v][j] * dp[u][k - j - 1];
        for (int j = 1; j <= k; j++)
            dp[u][j] += dp[v][j - 1];
    }
}

int main()
{
    int n, k;
    while (~scanf("%d%d", &n, &k))
    {
        init();

        for (int i = 0; i < n - 1; i++)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            addedge(u, v);
            addedge(v, u);
        }

        ans = 0;
        dfs(1, 0, k);
        cout << ans << endl;
    }
    return 0;
}