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  • 05-树6. Path in a Heap (25) 小根堆

    05-树6. Path in a Heap (25)

    Time Limit: 20 Sec  Memory Limit: 256 MB

    题目连接

    http://www.patest.cn/contests/mooc-ds2015spring/05-%E6%A0%916

    Description

    Insert a sequence of given numbers into an initially empty min-heap H. Then for any given index i, you are supposed to print the path from H[i] to the root.

    Input

    Each input file contains one test case. For each case, the first line gives two positive integers N and M (<=1000) which are the size of the input sequence, and the number of indices to be checked, respectively. Given in the next line are the N integers in [-10000, 10000] which are supposed to be inserted into an initially empty min-heap. Finally in the last line, M indices are given.

    Output

    For each index i in the input, print in one line the numbers visited along the path from H[i] to the root of the heap. The numbers are separated by a space, and there must be no extra space at the end of the line.

    Sample Input

    5 3 46 23 26 24 10 5 4 3

    Sample Output

    24 23 10 46 23 10 26 10

    HINT

    题意

     

    题解:

    随便拿个数组模拟一下小根堆就好了……

    代码:

    //qscqesze
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <set>
    #include <vector>
    #include <sstream>
    #include <queue>
    #include <typeinfo>
    #include <fstream>
    #include <map>
    #include <stack>
    typedef long long ll;
    using namespace std;
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    #define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
    #define test freopen("test.txt","r",stdin)  
    #define maxn 200001
    #define mod 10007
    #define eps 1e-9
    int Num;
    char CH[20];
    //const int inf=0x7fffffff;   //нчоч╢С
    const int inf=0x3f3f3f3f;
    /*
    
    inline void P(int x)
    {
        Num=0;if(!x){putchar('0');puts("");return;}
        while(x>0)CH[++Num]=x%10,x/=10;
        while(Num)putchar(CH[Num--]+48);
        puts("");
    }
    */
    inline ll read()
    {
        ll x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    inline void P(int x)
    {
        Num=0;if(!x){putchar('0');puts("");return;}
        while(x>0)CH[++Num]=x%10,x/=10;
        while(Num)putchar(CH[Num--]+48);
        puts("");
    }
    //**************************************************************************************
    int a[maxn];
    int main()
    {
        //test;
        int n=read(),m=read();
        for(int i=1;i<=n;i++)
        {
            a[i]=read();
            int tmp=i;
            while(tmp!=1&&a[tmp]<a[tmp/2])
            {
                swap(a[tmp],a[tmp/2]);
                tmp/=2;
            }
        }
        for(int i=1;i<=m;i++)
        {
            int first=1;
            int x=read();
            while(x!=0)
            {
                if(first)
                    printf("%d",a[x]),first=0;
                else
                    printf(" %d",a[x]);
                x/=2;
            }
            printf("
    ");
        }
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/4528942.html
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