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P2234 [HNOI2002]营业额统计

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 2e5 + 5;
const int M = 2e4 + 5;
const double eps = 1e-10;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e8
#define dbg(ax) cout << "now this num is " << ax << endl;
inline int read() {
    int f = 1;int x = 0;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
    return x*f;
}
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}


int f[N],cnt[N],value[N],sons[N][2],sz[N],allsz,rt;

inline bool get_which(int x) {return sons[f[x]][1] == x;}
inline void update(int x) {
    if(x) {
        sz[x] = cnt[x];
        if(sons[x][0]) sz[x] += sz[sons[x][0]];
        if(sons[x][1]) sz[x] += sz[sons[x][1]];
    }
    return ;
}
inline void Clear(int x) {sons[x][0] = sons[x][1] = f[x] = sz[x] = cnt[x] = value[x] = 0;}
inline void rotate(int x) {//旋转
    int fr = f[x],g_fr = f[fr],w_son = get_which(x);
    sons[fr][w_son] = sons[x][w_son ^ 1];
    f[sons[fr][w_son]] = fr;
    sons[x][w_son ^ 1] = fr;
    f[fr] = x;
    f[x] = g_fr;
    if(g_fr) sons[g_fr][sons[g_fr][1] == fr] = x;  
    update(fr);
    update(x);
}
inline void splay(int x) {
    for(int i;i = f[x];rotate(x)) 
        if(f[i]) rotate((get_which(x) == get_which(i)) ? i : x);
    rt = x;
}
inline void insert(int x) {
    if(!rt) {
        rt = ++allsz;
        sons[allsz][0] = sons[allsz][1] = f[allsz] = 0;
        sz[allsz] = cnt[allsz]++;
        value[allsz] = x;
        return ;
    }
    int now = rt,fa = 0;
    while(1) {
        if(x == value[now]) {
            cnt[now]++;
            update(now);
            update(fa);
            splay(now);
            break;
        }
        fa = now;
        now = sons[now][value[now] < x];
        if(!now) {
            f[++allsz] = fa;
            sons[allsz][0] = sons[allsz][1] = 0;
            sz[allsz] = cnt[allsz] = 1;
            sons[fa][value[fa] < x] = allsz;
            value[allsz] = x;
            update(fa);
            splay(allsz);
            break;
        }
    }
}
inline int find_num(int x) {//查询排名为x的数
    int now = rt;
    while(1) {
        if(sons[now][0] && x <= sz[sons[now][0]]) now = sons[now][0];
        else {
            int temp = (sons[now][0] ? sz[sons[now][0]] : 0) + cnt[now];
            if(x <= temp) return value[now];
            x -= temp;
            now = sons[now][1];
        }
    }
}
inline int find_rank(int x) {//查询x的排名
    int now = rt,ans = 0;
    while(1) {
        if(x < value[now]) now = sons[now][0];
        else {
            ans += (sons[now][0] ? sz[sons[now][0]] : 0);
            if(x == value[now]) {splay(now);return ans + 1;}
            ans += cnt[now];
            now = sons[now][1];
        }
    }
}
inline int find_pre() {
    int now = sons[rt][0];
    while(sons[now][1]) now = sons[now][1];
    return now;
}
inline int find_suffix() {
    int now = sons[rt][1];
    while(sons[now][0]) now = sons[now][0];
    return now;
}
inline void Delete(int x) {
    int h = find_rank(x);
    if(cnt[rt] > 1) {cnt[rt]--;update(rt);return;}
    if(!sons[rt][0] && !sons[rt][1]) {Clear(rt);rt = 0;return;}
    if(!sons[rt][0]) {
        int old_rt = rt;
        rt = sons[rt][1];
        f[rt] = 0;
        Clear(old_rt);
        return;
    }
    else if(!sons[rt][1]) {
        int old_rt = rt;
        rt = sons[rt][0];
        f[rt] = 0;
        Clear(old_rt);
        return;
    }
    int lemx = find_pre(),old_rt = rt;
    splay(lemx);
    sons[rt][1] = sons[old_rt][1];
    f[sons[old_rt][1]] = rt;
    Clear(old_rt);
    update(rt);
}
map<int,int> mp;
void solve() {
    int n;n = read();
    LL ans = 0;
    value[0] = INF;
    for(int i = 1;i <= n;++i) {
        int x;x = read();
        insert(x);
        if(i == 1) ans += x;
        else {
            if(mp[x] != 0) continue;
            int pre = value[find_pre()];
            int nxt = value[find_suffix()];
          //  printf("pre is %d nxt is %d\n",pre,nxt);
            int ma = min(abs(x - pre),abs(x - nxt));
            ans += ma;
        }
        mp[x]++;
    }
    printf("%lld\n",ans);
}
int main() {
    solve();
    system("pause");
    return 0;
}

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标签：rt,old,int,汇总,long,sons,const,数据结构
来源： https://www.cnblogs.com/zwjzwj/p/15337801.html