zoj 2334 分类： zoj templates 2015-08-11 23:12 2人阅读 评论(0) 收藏

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可并堆，左偏树，这是神犇“顺手学的东西”

以下是左偏树的合并操作代码。

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int merge(int x,int y)
{
//p[i] 结点i的权值，这里是维护大根堆
//d[i] 在i的子树中，i到叶子结点的最远距离.

    if(!x) return y;
    if(!y) return x;

    if(p[x] < p[y]) std::swap(x, y);

    r[x] = merge(r[x], y);
    if(r[x]) fa[r[x]] = x;

    if(d[l[x]] < d[r[x]]) std::swap(l[x], r[x]);//调整树的结构，使其满足左偏性质

    d[x] = d[r[x]] + 1;
    return x;
}

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这个题还是比较直白的。。。

时间复杂度：O(m∗log2n)

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#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <string>
#include <map>
#include <vector>
#include <stack>
#include <queue>
#include <utility>
#include <iostream>
#include <algorithm>

template<class Num>void read(Num &x)
{
    char c; int flag = 1;
    while((c = getchar()) < '0' || c > '9')
        if(c == '-') flag *= -1;
    x = c - '0';
    while((c = getchar()) >= '0' && c <= '9')
        x = (x<<3) + (x<<1) + (c-'0');
    x *= flag;
    return;
}
template<class Num>void write(Num x)
{
    if(x < 0) putchar('-'), x = -x;
    static char s[20];int sl = 0;
    while(x) s[sl++] = x%10 + '0',x /= 10;
    if(!sl) {putchar('0');return;}
    while(sl) putchar(s[--sl]);
}

const int maxn = 1e5 + 50 , maxm = maxn, Nya = -1;

int n, m, p[maxn];
int l[maxn], r[maxn], fa[maxn], d[maxn];

void init()
{
    memset(l, 0, sizeof(l));
    memset(r, 0, sizeof(r));
    memset(fa, 0, sizeof(fa));
    memset(d, 0, sizeof(d));

    for(int i = 1; i <= n; i++) read(p[i]);

    read(m);
}

int merge(int x,int y)
{
    if(!x) return y;
    if(!y) return x;

    if(p[x] < p[y]) std::swap(x, y);

    r[x] = merge(r[x], y);
    if(r[x]) fa[r[x]] = x;

    if(d[l[x]] < d[r[x]]) std::swap(l[x], r[x]);

    d[x] = d[r[x]] + 1;
    return x;
}
int root(int x)
{
    while(fa[x]) x = fa[x]; return x;
}
int dec(int t)
{
    int L = l[t], R = r[t];

    fa[L] = fa[R] = 0;
    l[t] = r[t] = d[t] = 0;

    p[t] >>= 1;

    return merge(merge(L, R), t);
}

void solve()
{
    int u, v, ans;

    d[0] = Nya;

    while(m--)
    {
        read(u), read(v);

        u = root(u), v = root(v);
        ans = (u == v) ? Nya : p[merge(dec(u), dec(v))];

        write(ans), puts("");       
    }
}

int main()
{
#ifndef ONLINE_JUDGE    
    freopen("2334.in","r",stdin);
    freopen("2334.out","w",stdout);
#endif

    while(scanf("%d", &n) != EOF)
        init(), solve();


#ifndef ONLINE_JUDGE    
    fclose(stdin);
    fclose(stdout);
#endif  
    return 0;
}

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