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  • hdu 5593 ZYB's Tree 树形dp

    ZYB's Tree

    Time Limit: 20 Sec

    Memory Limit: 256 MB

    题目连接

    http://acm.hdu.edu.cn/showproblem.php?pid=5593

    Description

    ZYB has a tree with N nodes,now he wants you to solve the numbers of nodes distanced no more than K for each node.
    the distance between two nodes(x,y) is defined the number of edges on their shortest path in the tree.

    To save the time of reading and printing,we use the following way:

    For reading:we have two numbers A and B,let fai be the father of node i,fa1=0,fai=(A∗i+B)%(i−1)+1 for i∈[2,N] .

    For printing:let ansi be the answer of node i,you only need to print the xor sum of all ansi.

    Input

    In the first line there is the number of testcases T.

    For each teatcase:

    In the first line there are four numbers N,K,A,B

    1≤T≤5,1≤N≤500000,1≤K≤10,1≤A,B≤1000000

    Output

    For T lines,each line print the ans.

    Please open the stack by yourself.

    N≥100000 are only for two tests finally.

    Sample Input

    1
    3 1 1 1

    Sample Output

    3

    HINT

    题意

     

    题解:

    我们维护两个dp,dp1[i][j]表示以i为根节点的子树,有多少个点在距离为j的位置

    dp2[i][j]表示并不是在子树的,有多少个离他距离为j的点

    首先很容易的在树上维护处dp1[i][j]

    然后我们再扫一遍用容斥做出dp2就好了

    代码:

    #include<iostream>
    #include<stdio.h>
    #include<vector>
    #include<cstring>
    using namespace std;
    #define maxn 500050
    vector<int> E[maxn];
    int dp[maxn][12];
    int dp2[maxn][12];
    
    int n,k,A,B;
    void dfs(int x)
    {
        dp[x][0]=1;
        for(int i=0;i<E[x].size();i++)
        {
            dfs(E[x][i]);
            for(int j=0;j<k;j++)
                dp[x][j+1]+=dp[E[x][i]][j];
        }
    }
    int main()
    {
        int t;scanf("%d",&t);
        for(int cas=1;cas<=t;cas++)
        {
            memset(dp,0,sizeof(dp));
            scanf("%d%d%d%d",&n,&k,&A,&B);
            for(int i=0;i<=n;i++)
                E[i].clear();
            for(int i=2;i<=n;i++)
            {
                int x = i;
                int y = (1LL*A+B*1LL)%(i*1LL-1)+1;
                E[y].push_back(x);
            }
            dfs(1);
            int Ans = 0;
            for(int x=1;x<=n;x++)
            {
                for(int i=0;i<E[x].size();i++)
                {
                    int y = E[x][i];
                    for(int j=0;j<=k;j++)
                        dp2[y][j]=0;
                    dp2[y][1]=dp[x][0];
                    for(int j=1;j<k;j++)
                        dp2[y][j+1]=dp[x][j]-dp[y][j-1]+dp2[x][j];
                }
                int now = 0;
                for(int i=0;i<=k;i++)
                    now+=dp[x][i]+dp2[x][i];
                Ans^=now;
            }
            printf("%d
    ",Ans);
        }
    
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/5023387.html
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