• Codeforces Round #277 (Div. 2) D. Valid Sets 暴力


    D. Valid Sets

    Time Limit: 20 Sec

    Memory Limit: 256 MB

    题目连接

    http://codeforces.com/contest/486/problem/D

    Description

    As you know, an undirected connected graph with n nodes and n - 1 edges is called a tree. You are given an integer d and a tree consisting of n nodes. Each node i has a value ai associated with it.

    We call a set S of tree nodes valid if following conditions are satisfied:

        S is non-empty.
        S is connected. In other words, if nodes u and v are in S, then all nodes lying on the simple path between u and v should also be presented in S.
        .

    Your task is to count the number of valid sets. Since the result can be very large, you must print its remainder modulo 1000000007 (109 + 7).

    Input

    The first line contains two space-separated integers d (0 ≤ d ≤ 2000) and n (1 ≤ n ≤ 2000).

    The second line contains n space-separated positive integers a1, a2, ..., an(1 ≤ ai ≤ 2000).

    Then the next n - 1 line each contain pair of integers u and v (1 ≤ u, v ≤ n) denoting that there is an edge between u and v. It is guaranteed that these edges form a tree.

    Output

    Print the number of valid sets modulo 1000000007.

    Sample Input

    1 4
    2 1 3 2
    1 2
    1 3
    3 4

    Sample Output

    8

    HINT

    题意

    给你一颗树,然后让你找到里面有多少棵子树,满足子树里面最大点减去最小点的差小于等于d

    题解:

    直接枚举每一个点为根,然后开始dfs,都假设这个根节点是这棵子树的最大值

    然后乘法原理搞一搞就好啦

    代码

    #include<iostream>
    #include<stdio.h>
    #include<vector>
    using namespace std;
    #define maxn 2005
    const int mod = 1e9 + 7;
    int a[maxn];
    int d,n;
    vector<int> G[maxn];
    long long dfs(int x,int pre,int k)
    {
        long long ans = 1;
        for(int i=0;i<G[x].size();i++)
        {
            int v = G[x][i];
            if(v == pre || (a[k] == a[v]&&v > k))continue;
            if(a[k]>=a[v]&&a[k]-a[v]<=d)
            {
                ans *= ( dfs(v,x,k) + 1 );
                while(ans>=mod)ans%=mod;
            }
        }
        return ans;
    }
    int main()
    {
        scanf("%d%d",&d,&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<n;i++)
        {
            int u,v;scanf("%d%d",&u,&v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        long long ans = 0;
        for(int i=1;i<=n;i++)
        {
            ans += dfs(i,-1,i);
            while(ans>=mod)ans%=mod;
        }
        printf("%lld
    ",ans);
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/4908658.html
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