HDU-6356 Glad You Came 线段树 ST表

HDU-6356

题意：有m次操作，每次操作通过给定的随机函数生成 l , r , v,使得在 l 到 r 区间内，所有的a【i】变为max（a[i] , v).

　　最后输出n个a【i】* i的异或和。

　

思路：线段树操作，每次维护区间的最小值，如果当前的v小于区间的最小值，直接return，lazy标记维护区间未加的最大值，必要时pushdown就ok；

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <list>
#include <cstdlib>
#include <iterator>
#include <cmath>
#include <iomanip>
#include <bitset>
#include <cctype>
using namespace std;
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#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 pair<ll ,ll > pll;
typedef pair<int ,int > pii;

//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 OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //

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;
}
// #define _DEBUG;         //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
/*-----------------------show time----------------------*/
                unsigned X,Y,Z,W;
                unsigned RNG61(){
                    X = X ^ (X<<);
                    X = X ^ (X>>);
                    X = X ^ (X<<);
                    X = X ^ (X>>);
                    W = X ^ (Y ^ Z);
                    X = Y;
                    Y = Z;
                    Z = W;
                    return Z;
                }

                const ll MOD = (<<);
                const int maxn = 1e5+;
                int n,m;

                ll sum[maxn*],lazy[maxn*];
                void build(int l,int r,int rt){
                    sum[rt] = ;
                    lazy[rt] = ;
                    if(l==r){
                        return;
                    }
                    int mid = (l + r)/;
                    build(l,mid,rt<<);
                    build(mid+,r,rt<<|);
                }
                void pushup(int rt){
                    sum[rt] = min(sum[rt<<] , sum[rt<<|]);
                }
                void pushdown(int rt){
                    if(lazy[rt]){
                        lazy[rt<<] = max(lazy[rt],lazy[rt<<]);
                        lazy[rt<<|] = max(lazy[rt],lazy[rt<<|]);
                        sum[rt<<] = max(sum[rt<<] , lazy[rt]);
                        sum[rt<<|] = max(sum[rt<<|] , lazy[rt]);
                        lazy[rt] = ;
                    }
                }
                void update(int l, int r, int rt,int L,int R,ll val){
                    if(sum[rt] > val)return;
                    if(l>=L && r <=R){

                        sum[rt] = max(sum[rt],val);

                        lazy[rt] = max(lazy[rt],val);
                        return;
                    }
                    pushdown(rt);
                    int mid = (l+r)/;
                    if(mid >= L)update(l,mid,rt<<,L,R,val);
                    if(mid < R)update(mid+,r, rt<<|, L,R,val);
                    pushup(rt);
                }
                ll ans = 0ll;
                void  g_ans(int l,int r,int rt,int L,int R){
                    if(l==r){
                            // debug(sum[rt]);
                            ans ^= (1ll*l*sum[rt]);
                            return;
                    }
                    pushdown(rt);
                    int mid = (l+r)/;
                    g_ans(l,mid,rt<<,L,R);
                    g_ans(mid+,r, rt<<|, L,R);
                }
int main(){
                int t;    scanf("%d", &t);
                while(t--){

                        scanf("%d%d", &n, &m);
                        scanf("%u%u%u", &X, &Y, &Z);
                        build(,n,);
                        // debug(X);
                        for(int i=; i<=m; i++){
                            ll s = RNG61();
                            ll t = RNG61();
                            int tmp1 = s%n+;
                            int tmp2 = t%n+;

                            int le = min(tmp1 , tmp2);
                            int ri = max(tmp1,  tmp2);
                            ll v = RNG61() % MOD;
                            update(,n,,le,ri,v);
                        }
                        ans = 0ll;
                        g_ans(,n,,,n);
                        printf("%lld\n", ans);
                }

            return ;
}

线段树

由于询问只有一次，就是最后的输出，所以可以用ST表，这道题算是一个逆向的构造ST表，推出dp【0】【i】的结果；

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <list>
#include <cstdlib>
#include <iterator>
#include <cmath>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <iostream>
using namespace std;
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#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 pair<ll ,ll > pll;
typedef pair<int ,int > pii;

//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 OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //

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;
}
// #define _DEBUG;         //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
/*-----------------------show time----------------------*/
                unsigned X,Y,Z,W;
                unsigned RNG61(){
                    X = X ^ (X<<);
                    X = X ^ (X>>);
                    X = X ^ (X<<);
                    X = X ^ (X>>);
                    W = X ^ (Y ^ Z);
                    X = Y;
                    Y = Z;
                    Z = W;
                    return Z;
                }

                const ll MOD = (<<);
                const int maxn = 1e5+;
                int n,m;

                ll dp[][maxn];
                int Log[maxn];
                void update(int le,int ri,ll val){
                    int k = Log[ri-le+];
                    dp[k][le] = max(dp[k][le], val);
                    dp[k][ri-(<<k)+] = max(val,dp[k][ri-(<<k)+]);
                }
                void solve(){
                    cin>>n>>m>>X>>Y>>Z;
                    for(int j=; j<;j++){
                        for(int i=; i<=n; i++){
                            dp[j][i] = ;
                        }
                    }

                    for(int i=; i<=m; i++){
                        int t1 = RNG61()%n + ;
                        int t2 = RNG61()%n + ;
                        int le = min(t1, t2);
                        int ri = max(t1, t2);
                        ll val = RNG61() % MOD;

                        update(le,ri,val);
                        // cout<<val<<endl;
                    }

                    for(int j=; j; j--){
                        for(int i=; i+(<<(j-)) <=n; i++){
                            dp[j-][i] = max(dp[j][i] , dp[j-][i]);
                            dp[j-][i+(<<(j-))] = max(dp[j-][i+(<<(j-))] ,dp[j][i]);
                        }
                    }

                    ll ans = ;
                    for(int i=; i<=n; i++){
                        ans = ans ^ (i * dp[][i]);
                        // cout<<dp[0][i]<<" ";
                    }
                    // cout<<endl;
                    cout<<ans<<endl;
                }

int main(){
            OKC;
            Log[] = ;
            for(int i=; i<maxn; i++){
                Log[i] = Log[i>>] + ;
            }
            int t;    cin>>t;

            while(t--){
                solve();
            }
            return ;
}

ST表