CF 551 E GukiZ and GukiZiana

Posted 2021-02-05 ckxkexing

https://codeforces.com/contest/551/problem/E

分块真强。

题意就是1、区间加，2、询问整个区间中，最远的两个x的距离。

分块，然后，每次找位子用二分找即可。

 

 

#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>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#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;
typedef pair<ll,int>pli;
//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 ‘
‘
//#define R register
#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)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

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;
}


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

        const int maxn = 5e5+9;
        ll a[maxn];
        int be[maxn];
        int blo;
        ll add[maxn];
        vector<pli>v[maxn];
        int n,m;
        void rebuild(int id){
            v[id].clear();

            for(int i=(id-1)*blo + 1; i<= min(n,id*blo); i++){
                a[i] += add[id];
                v[id].pb(pli(a[i],i));
            }
            sort(v[id].begin(),v[id].end());
            add[id] = 0;
        }
int main(){
        scanf("%d%d", &n, &m);
        for(int i=1; i<=n; i++) scanf("%lld", &a[i]);
        // blo = (int)sqrt(n);
        blo = 1000;
        for(int i=1; i<=n; i++){
            be[i] = (i-1)/blo + 1;
            v[be[i]].pb(pli(a[i],i));
        }
        for(int i=1; i<=be[n]; i++){
            sort(v[i].begin(),v[i].end());
        }

        while(m--){
            int op; scanf("%d", &op);
            if(op == 1){
                int l,r,x;
                scanf("%d%d%d", &l, &r, &x);
                for(int i=l; i<= min(r, be[l]*blo); i++){
                    a[i] += x;
                }
                rebuild(be[l]);

                if(be[l] < be[r]){
                    for(int i=be[l]+1; i<=be[r]-1; i++){
                        add[i] += x;
                    }
                
                     for(int i=(be[r]-1)*blo+1; i<= r; i++){
                          a[i] += x;
                     }
                     rebuild(be[r]);
                }
            }
            else {
                int x;  scanf("%d", &x);
                int le=n+1,ri=-1;
                for(int i=1; i<=be[n]; i++){
                    int tmp = lower_bound(v[i].begin(),v[i].end(),pli(1ll*(x-add[i]),0)) - v[i].begin();

                    if(v[i][tmp].fi == x-add[i]){
                        le = min(le, v[i][tmp].se);
                    }
                    tmp = upper_bound(v[i].begin(), v[i].end(), pli(1ll*(x-add[i]),n+1)) - v[i].begin() - 1;
                    if(tmp>=0 && v[i][tmp].fi == x-add[i]){
                        ri = max(ri, v[i][tmp].se);
                    }
                }
                if(le <= ri){
                    printf("%d
", ri - le);
                }   
                else puts("-1");

            }
        }
        return 0;
}

View Code

自己一开始把 块的id 和 i 搞混了...