【ZOJ2334】Monkey King

题目大意

有 n(n≤105) 只猴子，初始每只猴子为自己猴群的猴王，每只猴子有一个初始的力量值。这些猴子会有 m 次会面。每次两只猴子 x,y 会面，若 x,y 属于同一个猴群输出 −1 ，否则将 x,y 所在猴群的猴王的力量值减半，然后合并这两个猴群。新猴群中力量值最高的为猴王。输出新猴王的力量值。

简要分析

我们需要一种数据结构维护元素，支持快速合并，查找最大值。普通的堆并不能快速合并，所以大神们搞出了可并堆这种东西。
左偏树是一种比较好写的可并堆，详细请看BYVOID大神（黑科技：pbds库里有可并堆模板哦！）

AC code :

#include <set>#include <map>#include <ctime>#include <cmath>#include <queue>#include <cstdio>#include <cctype>#include <vector>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define MP make_pair#define PB push_backtypedef long long LL;typedef pair<int, int> pii;inline int read(){    int x = 0; int v = 1, c = getchar();    for(; c < '0' || c > '9'; c = getchar()) if(c == '-') v = -1;    for(;c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';    return x * v;}template<typename T>inline void write(T x, int ch = 10){    int s[20], t = 0;    if(x < 0) { x = -x; putchar('-'); }    do s[t++] = x % 10; while(x /= 10);    while(t--) putchar('0' + s[t]);    putchar(ch);}const int maxn = 100005;int n, m, v[maxn];int lc[maxn], rc[maxn], d[maxn], fa[maxn];void input(){    for(int i = 1; i <= n; ++i)        v[i] = read();    m = read();}inline void clear(const int &x){    d[x] = lc[x] = rc[x] = 0;}inline void reset(const int &x){    fa[x] = x;}void build(){    for(int i = 1; i <= n; ++i)    {        reset(i);        clear(i);    }}int getfa(const int &x){    return fa[x] == x ? x : fa[x] = getfa(fa[x]);}int merge(int a, int b){    if(a == 0) return b;    if(b == 0) return a;    if(v[a] < v[b]) swap(a, b);    rc[a] = merge(rc[a], b);    if(rc[a]) fa[rc[a]] = a;    if(d[lc[a]] < d[rc[a]]) swap(lc[a], rc[a]);    d[a] = rc[a] ? d[rc[a]] + 1 : 0;    return a;}int pop(int x){    int l = lc[x], r = rc[x];    reset(l); reset(r);    reset(x); clear(x);    return merge(l, r);}int modify(int x){    v[x] >>= 1;    return merge(x, pop(x));}void solve(){    register int x, y, z;    for(int i = 1; i <= m; ++i)    {        x = getfa(read()); y = getfa(read());        if(x == y) puts("-1");        else        {            x = modify(x);            y = modify(y);            z = merge(x, y);            write(v[z]);        }    }}int main(){#ifndef ONLINE_JUDGE    freopen("input.txt", "r", stdin);    freopen("output.txt", "w", stdout);#endif    while(~scanf("%d", &n))    {        input();        build();        solve();    }#ifndef ONLINE_JUDGE    fprintf(stderr, "Run Time: %dms\n", clock());    fclose(stdin), fclose(stdout);#endif    return 0;}

题外话(黑科技)

使用pbds中的可并堆

一点介绍

原文在这里
包含：ext/pb_ds/priority_queue.hpp
声明：__gnu_pbds::priority_queue
模板参数：

template < typename Value_Type ,    typename Cmp_Fn = std :: less < Value_Type >,    typename Tag = pairing_heap_tag ,    typename Allocator = std :: allocator <char > >class priority_queue

Value_Type：类型
Cmp_Fn：自定义比较器
Tag：堆的类型。可以是

binary_heap_tag（二叉堆）binomial_heap_tag（二项堆）rc_binomial_heap_tag pairing_heap_tag（配对堆）thin_heap_tag

Allocator：不用管

使用

相比STL中的priority_queue，可以

用begin()和end()获取迭代器从而遍历
删除单个元素

void erase(point_iterator)

增加或减少某一元素的值

void modify(point_iterator, const_reference)

合并，把other合并到*this，并把other清空

void join(priority_queue &other)

性能分析

五种操作：push、pop、modify、erase、join

pairing_heap_tag：push和join O(1)，其余均摊 O(logn)
binary_heap_tag：只支持push和pop，均为均摊 O(logn)
binomial_heap_tag：push为均摊 O(1)，其余为 O(logn)
rc_binomial_heap_tag：push为 O(1)，其余为 O(logn)
thin_heap_tag：push为 O(1)，不支持join，其余为 O(logn)；但是如果只有increase_key，modify均摊 O(1)
不支持不是不能用，而是用起来很慢

黑科技演示

#include <cstdio>#include <cstring>#include <algorithm>#include <ext/pb_ds/priority_queue.hpp>using namespace __gnu_pbds;inline int read(){    int x = 0; int v = 1, c = getchar();    for(; c < '0' || c > '9'; c = getchar()) if(c == '-') v = -1;    for(;c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';    return x * v;}template<typename T>inline void write(T x, int ch = 10){    int s[20], t = 0;    if(x < 0) { x = -x; putchar('-'); }    do s[t++] = x % 10; while(x /= 10);    while(t--) putchar('0' + s[t]);    putchar(ch);}const int maxn = 100005;int n, m, v[maxn];int fa[maxn];priority_queue<int> q[maxn];int getfa(const int &x){    return fa[x] == x ? x : fa[x] = getfa(fa[x]);}void input(){    for(int i = 1; i <= n; ++i)    {        fa[i] = i;        v[i] = read();        q[i].clear();        q[i].push(v[i]);    }    m = read();}void solve(){    register int x, y, z;    for(int i = 1; i <= m; ++i)    {        x = getfa(read()); y = getfa(read());        if(x == y) puts("-1");        else        {            z = q[x].top();            q[x].pop();            q[x].push(z >> 1);            z = q[y].top();            q[y].pop();            q[y].push(z >> 1);            fa[y] = x;            q[x].join(q[y]);            write(q[x].top());        }    }}int main(){#ifndef ONLINE_JUDGE    freopen("input.txt", "r", stdin);    freopen("output.txt", "w", stdout);#endif    while(~scanf("%d", &n))    {        input();        solve();    }#ifndef ONLINE_JUDGE    fclose(stdin), fclose(stdout);#endif    return 0;}