Codeforces 551E - GukiZ and GukiZiana(分块)

Problem E. GukiZ and GukiZiana

Solution：

　　先分成N=sqrt（n）块，然后对这N块进行排序。

     利用二分查找确定最前面和最后面的位置。

　　

#include <bits/stdc++.h>



using namespace std;



typedef long long ll;

vector<int> s[1000];

ll add[1000], a[512346], pos[512346];

ll n, q, bk, N;



bool cmp( int x, int y )

{

    if( a[x] == a[y]  ) return x < y;

    return a[x] < a[y];

}



inline void modify( ll l, ll r, ll x )

{

    int k = pos[l], t = pos[r];



    if( k == t ) {

        for( ll i = l; i <= r; ++i )

            a[i] += x;

        sort( s[k].begin(), s[k].end(), cmp );

        return ;

    }



    for( ll i = k + ( pos[l - 1] == k ); i <= t - ( pos[r + 1] == t ) ; ++i )

        add[i] += x;



    if( pos[l - 1] == k ) {

        for( ll i = l; pos[i] == k; ++i ) {

            a[i] += x;

        }

        sort( s[k].begin(), s[k].end(), cmp );

    }



    if( pos[r + 1] == t ) {

        for( ll i = r; pos[i] == t; --i ) {

            a[i] += x;

        }

        sort( s[t].begin(), s[t].end(), cmp );

    }

}



inline ll query( ll x )

{

    int l = -1, r = -1, i;

    for( i = 1; i <= N; ++i ) {

        a[0] = x - add[i];

        auto it = lower_bound( s[i].begin(), s[i].end(), 0, cmp );

        if( it == s[i].end() ) continue;

        if( a[*it] + add[i] == x ) {

            l = *it;

            break;

        }

    }



    if( l == -1 ) return l;

    for( int  j = N; j >= i; --j ) {

        a[n + 1] = x - add[j];

        auto it = lower_bound( s[j].begin(), s[j].end(), n + 1, cmp );

        if( it == s[j].begin() ) continue;

        it--;

        if( a[*it] + add[j] == x ) {

            r = *it;

            break;

        }

    }

    return r - l;

}



int main()

{

    ios::sync_with_stdio( 0 );

    cin >> n >> q;

    bk = ceil( sqrt( 1.*n ) + 0.005 );

    for( int i = 1; i <= n; ++i ) {

        cin >> a[i];

        pos[i] = ( i - 1 ) / bk + 1;

        s[pos[i]].push_back( i );

    }

    N = ( n - 1 ) / bk + 1;

    for( int i = 1; i <= N; ++i ) {

        sort( s[i].begin(), s[i].end(), cmp );

    }

    ll cmd, l, r, x;

    for( int i = 1; i <= q; ++i ) {

        cin >> cmd;

        if( cmd == 1 ) {

            cin >> l >> r >> x;

            modify( l, r, x );

        } else {

            cin >> x;

            cout << query( x ) << "\n";

        }

    }

}

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