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  • CF 551 E GukiZ and GukiZiana

    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... 

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  • 原文地址:https://www.cnblogs.com/ckxkexing/p/10307564.html
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