gym/102091

https://codeforces.com/gym/102091

2018-2019 ACM-ICPC, Asia Nakhon Pathom Regional Contest

 A Flying Squirrel

题意

有n个柱子。m次询问。

每次询问从x号柱子跳到y号柱子，最多能踩几个柱子。

每次跳跃只能向低的柱子跳，且中间不能有高于起跳点的柱子。

思路

化数列为DAG，对于一个柱子u来说，向左跳到能跳的区域中最高的柱子v，我们连边u->v。然后就类似树上的操作。

注意dfs和bfs中都要做好打标记的操作。

 

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define debug(x) cerr<<#x << " := " << x << endl;
#define bug cerr<<"-----------------------"<<endl;
#define FOR(a, b, c) for(int a = b; a <= c; ++ a)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;


template<class T> void _R(T &x) { cin >> x; }
void _R(int &x) { scanf("%d", &x); }
void _R(ll &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template<class T, class... U> void R(T &head, U &... tail) { _R(head); R(tail...); }


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}

const int inf = 0x3f3f3f3f;

const int mod = 1e9+7;

/**********showtime************/
            const int maxn = 1e5+9;
            int a[maxn];
            vector<int>mp[maxn];
            vector<int>hi[maxn];

            int mx[maxn<<2];
            void build(int le, int ri, int rt){
                if(le == ri) {
                    mx[rt] = a[le];
                    return;
                }
                int mid = (le + ri) >> 1;
                build(le, mid, rt<<1);
                build(mid+1,ri,rt<<1|1);
                mx[rt] = max(mx[rt<<1], mx[rt<<1|1]);
            }
            int query(int L, int R, int le, int ri, int rt){
                if(le >= L && ri <= R) {
                    return mx[rt];
                }
                int mid = (le + ri) >> 1;
                int res = 0;
                if(mid >= L) res = max(res, query(L, R, le, mid, rt<<1));
                if(mid < R) res = max(res, query(L, R, mid+1, ri, rt<<1|1));
                return res;
            }
            int low[maxn],up[maxn];
            int st[maxn];
            int top = 0;
            int dp[maxn], mdp[maxn];
            int vis[maxn],used[maxn];
            void dfs(int u, int o) {
                vis[u] = true;
                dp[u] = dp[o] + 1;
                mdp[u] = 0;
                for(int v : mp[u]) {
                    if(!vis[v]) dfs(v, u);
                    mdp[u] = max(mdp[u], mdp[v] + 1);
                }
            }
            bool check(int x, int y) {
                if(y <= up[x] && y >= low[x]) return true;
                if(x <= up[y] && x >= low[y]) return true;
                return false;
            }
int main(){
            int n,m;
            scanf("%d%d", &n, &m);
            for(int i=1; i<=n; i++) {
                scanf("%d", &a[i]);
                hi[a[i]].pb(i);
            }

            for(int i=1; i<=n; i++) {
                while(top > 0 && a[st[top]] < a[i]) top--;
                if(top == 0) low[i] = 1;
                else low[i] = st[top] + 1;
                st[++top] = i;
            }

            top = 0;
            for(int i=n; i>=1; i--) {
                while(top > 0 && a[st[top]] < a[i])top--;
                if(top == 0) up[i] = n;
                else up[i] = st[top] - 1;
                st[++top] = i;
            }

            build(1, n, 1);
            int big = query(1, n, 1, n, 1);

            int root = n + 1;
            queue<int>que;
            for(int v : hi[big]) {
                mp[root].pb(v);
                que.push(v);
            }

            while(!que.empty()) {
                int u = que.front();    que.pop();
                int le = low[u], ri = up[u];
                if(le < u) {
                    int mx = query(le, u - 1, 1, n, 1);
                    int id = lower_bound(hi[mx].begin(), hi[mx].end(), le) - hi[mx].begin();

                    for(int i=id; i<hi[mx].size(); i++) {
                        if(hi[mx][i] >= u) break;
                        int v = hi[mx][i];
                        mp[u].pb(v);
                        if(used[v] == 0)
                        {
                            que.push(v);
                            used[v] = 1;
                        }
                    }
                }
                if(u < ri) {
                    int mx = query(u+1, ri, 1, n, 1);
                    int id = lower_bound(hi[mx].begin(), hi[mx].end(), u+1) - hi[mx].begin();
                    for(int i=id; i<hi[mx].size(); i++) {
                        if(hi[mx][i] > ri) break;
                        int v = hi[mx][i];
                        mp[u].pb(v);
                        if(used[v] == 0) {
                            used[v] = 1;
                            que.push(v);
                        }
                    }
                }
            }

            dfs(root, root);

            for(int i=1; i<=m; i++) {
                int x,y;
                scanf("%d%d", &x, &y);
                if(y == 0) {
                    printf("%d\n", mdp[x]);
                }
                else {
                    if(!check(x, y)) puts("0");
                    else printf("%d\n", abs(dp[x] - dp[y]));
                }
            }
            return 0;
}

 

E How Many Groups

 

K The Stream of Corning 2

题意

有n个事物会出现在河流中，每个事物会出现于le到ri秒。问第i秒第K小是多少。保证le和i在出现顺序中是递增的。

思路

先离散化，然后开一个树状数组，出现了的事物就插入权值树状数组，并记下消失的时间ri，等查询操作前把消失的事物清除。 

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

/*
        
⊂_ヽ
　 ＼＼ Λ＿Λ  来了老弟
　　 ＼('ㅅ')
　　　 >　⌒ヽ
　　　/ 　 へ＼
　　 /　　/　＼＼
　　 ﾚ　ノ　　 ヽ_つ
　　/　/
　 /　/|
　(　(ヽ
　|　|、＼
　| 丿 ＼ ⌒)
　| |　　) /
'ノ )　　Lﾉ

*/

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost ios::sync_with_stdio(false);cin.tie(0)
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);


const ll oo = 1ll<<17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}
struct FastIO {
    static const int S = 4e6;
    int wpos;
    char wbuf[S];
    FastIO() : wpos(0) {}
    inline int xchar() {
        static char buf[S];
        static int len = 0, pos = 0;
        if (pos == len)
            pos = 0, len = fread(buf, 1, S, stdin);
        if (pos == len) exit(0);
        return buf[pos++];
    }
    inline int xuint() {
        int c = xchar(), x = 0;
        while (c <= 32) c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
        return x;
    }
    inline int xint()
    {
        int s = 1, c = xchar(), x = 0;
        while (c <= 32) c = xchar();
        if (c == '-') s = -1, c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
        return x * s;
    }
    inline void xstring(char *s)
    {
        int c = xchar();
        while (c <= 32) c = xchar();
        for (; c > 32; c = xchar()) * s++ = c;
        *s = 0;
    }
    inline void wchar(int x)
    {
        if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;
        wbuf[wpos++] = x;
    }
    inline void wint(int x)
    {
        if (x < 0) wchar('-'), x = -x;
        char s[24];
        int n = 0;
        while (x || !n) s[n++] = '0' + x % 10, x /= 10;
        while (n--) wchar(s[n]);
        wchar('\n');
    }
    inline void wstring(const char *s)
    {
        while (*s) wchar(*s++);
    }
    ~FastIO()
    {
        if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;
    }
} io;   
inline void cmax(int &x,int y){if(x<y)x=y;}
inline void cmax(ll &x,ll y){if(x<y)x=y;}
inline void cmin(int &x,int y){if(x>y)x=y;}
inline void cmin(ll &x,ll y){if(x>y)x=y;}

/*-----------------------showtime----------------------*/

            const int maxn = 1e6+9;
            vector<int>v;
            int getid(int x){
                return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
            }
            struct ask
            {
                int op;
                int a,b,c;
            }e[maxn];
            int sum[maxn];
            int lowbit(int x){
                return x & (-x);
            }
            void add(int x,int c){
                while(x < maxn){
                    sum[x] += c;
                    x = x + lowbit(x);
                }
            }
            int cal(int x){
                int res = 0;
                while(x > 0){
                    res += sum[x];
                    x -= lowbit(x);
                }
                return res;
            }
            vector<int>er[maxn];

int main(){
            int T,n;  //scanf("%d", &T);
            T = io.xint();
            rep(cas, 1, T){
                printf("Case %d:\n", cas);
                // scanf("%d", &n);
                n = io.xint();
                v.clear();
                for(int i=0; i<maxn; i++) er[i].clear();
                memset(sum, 0, sizeof(sum));
                rep(i, 1, n) {
                    // scanf("%d", &e[i].op);
                    e[i].op = io.xint();
                    if(e[i].op == 1) {
                        // scanf("%d%d%d", &e[i].a, &e[i].b, &e[i].c);
                        e[i].a = io.xint(); e[i].b = io.xint(); e[i].c = io.xint();
                        v.pb(e[i].a),v.pb(e[i].c);
                    }
                    else {
                        // scanf("%d%d", &e[i].a, &e[i].b);
                        e[i].a = io.xint(); e[i].b = io.xint();
                        v.pb(e[i].a);
                    }
                }
                sort(v.begin(), v.end());
                v.erase(unique(v.begin(), v.end()), v.end());
                int la = 0;
                rep(i, 1, n){
                    
                    if(e[i].op == 1) {
                        add(e[i].b, 1);
                        int t = getid(e[i].c);
                        er[t+1].pb(e[i].b);
                    }
                    else {
                        for(int k=la; k <= getid(e[i].a); k++)
                            for(int j=0; j<er[k].size(); j++){
                                add(er[k][j], -1);
                            }
                        la = getid(e[i].a)+1;
                        int res = -1, le = 1, ri = maxn-1;  

                        while(le <= ri){
                            int mid = (le + ri) >> 1;
                            if(cal(mid) >= e[i].b) {res = mid, ri = mid - 1;}
                            else le = mid + 1;
                        }
                        if(res == -1) puts("-1");
                        else printf("%d\n", res);
                    }
                } 

            }

            return 0;
}