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=;int f=;char ch=getchar();

    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();

    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();

    return x=f?-x:x;

}

const int inf = 0x3f3f3f3f;

const int mod = 1e9+;

/**********showtime************/

            const int maxn = 1e5+;

            int a[maxn];

            vector<int>mp[maxn];

            vector<int>hi[maxn];

            int mx[maxn<<];

            void build(int le, int ri, int rt){

                if(le == ri) {

                    mx[rt] = a[le];

                    return;

                }

                int mid = (le + ri) >> ;

                build(le, mid, rt<<);

                build(mid+,ri,rt<<|);

                mx[rt] = max(mx[rt<<], mx[rt<<|]);

            }

            int query(int L, int R, int le, int ri, int rt){

                if(le >= L && ri <= R) {

                    return mx[rt];

                }

                int mid = (le + ri) >> ;

                int res = ;

                if(mid >= L) res = max(res, query(L, R, le, mid, rt<<));

                if(mid < R) res = max(res, query(L, R, mid+, ri, rt<<|));

                return res;

            }

            int low[maxn],up[maxn];

            int st[maxn];

            int top = ;

            int dp[maxn], mdp[maxn];

            int vis[maxn],used[maxn];

            void dfs(int u, int o) {

                vis[u] = true;

                dp[u] = dp[o] + ;

                mdp[u] = ;

                for(int v : mp[u]) {

                    if(!vis[v]) dfs(v, u);

                    mdp[u] = max(mdp[u], mdp[v] + );

                }

            }

            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=; i<=n; i++) {

                scanf("%d", &a[i]);

                hi[a[i]].pb(i);

            }

            for(int i=; i<=n; i++) {

                while(top >  && a[st[top]] < a[i]) top--;

                if(top == ) low[i] = ;

                else low[i] = st[top] + ;

                st[++top] = i;

            }

            top = ;

            for(int i=n; i>=; i--) {

                while(top >  && a[st[top]] < a[i])top--;

                if(top == ) up[i] = n;

                else up[i] = st[top] - ;

                st[++top] = i;

            }

            build(, n, );

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

            int root = n + ;

            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 - , , n, );

                    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] == )

                        {

                            que.push(v);

                            used[v] = ;

                        }

                    }

                }

                if(u < ri) {

                    int mx = query(u+, ri, , n, );

                    int id = lower_bound(hi[mx].begin(), hi[mx].end(), u+) - 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] == ) {

                            used[v] = ;

                            que.push(v);

                        }

                    }

                }

            }

            dfs(root, root);

            for(int i=; i<=m; i++) {

                int x,y;

                scanf("%d%d", &x, &y);

                if(y == ) {

                    printf("%d\n", mdp[x]);

                }

                else {

                    if(!check(x, y)) puts("");

                    else printf("%d\n", abs(dp[x] - dp[y]));

                }

            }

            return ;

}

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<<;

const ll mos = 0x7FFFFFFF;  //

const ll nmos = 0x80000000;  //-2147483648

const int inf = 0x3f3f3f3f;

const ll inff = 0x3f3f3f3f3f3f3f3f; //

const int mod = 1e9+;

const double esp = 1e-;

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=;int f=;char ch=getchar();

    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();

    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();

    return x=f?-x:x;

}

struct FastIO {

    static const int S = 4e6;

    int wpos;

    char wbuf[S];

    FastIO() : wpos() {}

    inline int xchar() {

        static char buf[S];

        static int len = , pos = ;

        if (pos == len)

            pos = , len = fread(buf, , S, stdin);

        if (pos == len) exit();

        return buf[pos++];

    }

    inline int xuint() {

        int c = xchar(), x = ;

        while (c <= ) c = xchar();

        for (; '' <= c && c <= ''; c = xchar()) x = x *  + c - '';

        return x;

    }

    inline int xint()

    {

        int s = , c = xchar(), x = ;

        while (c <= ) c = xchar();

        if (c == '-') s = -, c = xchar();

        for (; '' <= c && c <= ''; c = xchar()) x = x *  + c - '';

        return x * s;

    }

    inline void xstring(char *s)

    {

        int c = xchar();

        while (c <= ) c = xchar();

        for (; c > ; c = xchar()) * s++ = c;

        *s = ;

    }

    inline void wchar(int x)

    {

        if (wpos == S) fwrite(wbuf, , S, stdout), wpos = ;

        wbuf[wpos++] = x;

    }

    inline void wint(int x)

    {

        if (x < ) wchar('-'), x = -x;

        char s[];

        int n = ;

        while (x || !n) s[n++] = '' + x % , x /= ;

        while (n--) wchar(s[n]);

        wchar('\n');

    }

    inline void wstring(const char *s)

    {

        while (*s) wchar(*s++);

    }

    ~FastIO()

    {

        if (wpos) fwrite(wbuf, , wpos, stdout), wpos = ;

    }

} 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+;

            vector<int>v;

            int getid(int x){

                return lower_bound(v.begin(), v.end(), x) - v.begin() + ;

            }

            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 = ;

                while(x > ){

                    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, , T){

                printf("Case %d:\n", cas);

                // scanf("%d", &n);

                n = io.xint();

                v.clear();

                for(int i=; i<maxn; i++) er[i].clear();

                memset(sum, , sizeof(sum));

                rep(i, , n) {

                    // scanf("%d", &e[i].op);

                    e[i].op = io.xint();

                    if(e[i].op == ) {

                        // 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 = ;

                rep(i, , n){

                    if(e[i].op == ) {

                        add(e[i].b, );

                        int t = getid(e[i].c);

                        er[t+].pb(e[i].b);

                    }

                    else {

                        for(int k=la; k <= getid(e[i].a); k++)

                            for(int j=; j<er[k].size(); j++){

                                add(er[k][j], -);

                            }

                        la = getid(e[i].a)+;

                        int res = -, le = , ri = maxn-;  

                        while(le <= ri){

                            int mid = (le + ri) >> ;

                            if(cal(mid) >= e[i].b) {res = mid, ri = mid - ;}

                            else le = mid + ;

                        }

                        if(res == -) puts("-1");

                        else printf("%d\n", res);

                    }

                } 

            }

            return ;

}

巴特西

gym/102091

A Flying Squirrel

# 题意

# 思路

E How Many Groups

K The Stream of Corning 2

# 题意

# 思路

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