2017.3.18 机房测试(线段树，树链剖分)

来源：互联网发布：华能辞职知乎编辑：程序博客网时间：2024/06/14 06:16

丁神远在上海依然不忘为我们出题。。。

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【解题报告】

代码如下：

#include <cstdio>#include <iostream>using namespace std;const int N = 1000000 + 10;int n;int aa[N + N];int bit[N * 24];long long ans, cur;void modify( int pos, int delta ) {    for( int i = pos; i <= n; i += i & -i )        bit[i] += delta;}int query( int rg ) {    int rt = 0;    for( int i = rg; i; i -= i & -i )        rt += bit[i];    return rt;}int main() {    freopen("rotinv.in","r",stdin);    freopen("rotinv.out","w",stdout);    scanf( "%d", &n );    for( int i = 1; i <= n; i++ ) {        scanf( "%d", aa + i );        aa[n + i] = aa[i];    }    for( int i = 1; i <= n; i++ ) {        cur += (i - 1) - query( aa[i] );        modify( aa[i], +1 );    }    for( int i = n + 1; i <= n + n; i++ ) {        modify( aa[i - n], -1 );        cur += n - 1 - query( aa[i] );        cur -= query( aa[i - n] - 1 );        modify( aa[i], +1 );        ans += cur;    }    cout << ans << endl;    return 0；}

这里写图片描述
【解题报告】

代码如下：

#include <cstdio>#include <iostream>using namespace std;const int N = 500000 + 10;struct Node {    int lf, rg;    int c1, c2, vmax;    Node *ls, *rs;    int query( int h )     {        if( lf == rg )         {            return h < vmax;        }         else         {            if( h < ls->vmax )                return ls->query(h) + c2;            else                return rs->query(h);        }    }    void update()     {        vmax = max( ls->vmax, rs->vmax );        c2 = rs->query( ls->vmax );        c1 = ls->c1 + c2;    }}pool[N*2], *tail = pool, *root;int n, m;int hh[N];Node *qu[N]; int bg, ed;Node *build( int lf, int rg ) {    Node *nd = ++tail;    nd->lf = lf;    nd->rg = rg;    if( lf == rg )     {        nd->c1 = 1;        nd->c2 = 0; //  no sense        nd->vmax = hh[lf];    }     else     {        int mid = (lf + rg) >> 1;        nd->ls = build( lf, mid );        nd->rs = build( mid+1, rg );        nd->update();    }    return nd;}void fetch( Node *nd, int L, int R ) {    if( L <= nd->lf && nd->rg <= R )     {        qu[++ed] = nd;        return;    }    int mid = (nd->lf + nd->rg) >> 1;    if( L <= mid )        fetch( nd->ls, L, R );    if( R > mid )        fetch( nd->rs, L, R );}int query( int L, int R ) {    bg = 1, ed = 0;    fetch( root, L, R );    int curh = 0;    int ans = 0;    for( int i = bg; i <= ed; i++ )     {        ans += qu[i]->query( curh );        curh = max( curh, qu[i]->vmax );    }    return ans;}int main() {    freopen("rise.in","r",stdin);    freopen("rise.out","w",stdout);    scanf( "%d%d", &n, &m );    for( int i = 1; i <= n; i++ )        scanf( "%d", hh + i );    root = build( 1, n );    while( m-- )     {        int L, R;        scanf( "%d%d", &L, &R );        printf( "%d\n", query(L,R) );    }}

这里写图片描述
【解题报告】

代码如下：

#include <cstdio>const int N = 100000 + 10;const int M = N + N;const int Mod = 31;struct Info {    int v, len;    Info(){}    Info( int v, int len ):v(v),len(len){}};struct Node {    Info ab, ba;    Node *ls, *rs;}pool[N*4], *tail = pool, *root;int n, m;int head[N], wght[M], dest[M], last[M], etot;int dep[N], siz[N], son[N], top[N], fat[N], val[N], in[N], rin[N], idc;int tpow[N];void init() {    tpow[0] = 1;    for( int i = 1; i < N; i++ )        tpow[i] = tpow[i-1] * 10 % Mod;}Info merge( const Info &a, const Info &b ) {    return Info( (a.v * tpow[b.len] + b.v) % Mod, a.len + b.len );}Node *build( int lf, int rg ) {    Node *nd = ++tail;    if( lf == rg ) {        nd->ab = nd->ba = Info( val[rin[lf]], 1 );    } else {        int mid = (lf + rg) >> 1;        nd->ls = build( lf, mid );        nd->rs = build( mid + 1, rg );        nd->ab = merge( nd->ls->ab, nd->rs->ab );        nd->ba = merge( nd->rs->ba, nd->ls->ba );    }    return nd;}Info query_ab( Node *nd, int lf, int rg, int L, int R ) {    if( L <= lf && rg <= R ) return nd->ab;    int mid = (lf + rg) >> 1;    if( R <= mid ) return query_ab( nd->ls, lf, mid, L, R );    else if( L > mid ) return query_ab( nd->rs, mid+1, rg, L, R );    else return merge(             query_ab( nd->ls, lf, mid, L, R ),             query_ab( nd->rs, mid+1, rg, L, R ) );}Info query_ba( Node *nd, int lf, int rg, int L, int R ) {    if( L <= lf && rg <= R ) return nd->ba;    int mid = (lf + rg) >> 1;    if( R <= mid ) return query_ba( nd->ls, lf, mid, L, R );    else if( L > mid ) return query_ba( nd->rs, mid+1, rg, L, R );    else return merge(            query_ba( nd->rs, mid+1, rg, L, R ),            query_ba( nd->ls, lf, mid, L, R ) );}void adde( int u, int v, int d ) {    etot++;    dest[etot] = v;    last[etot] = head[u];    wght[etot] = d;    head[u] = etot;}void dfs1( int u ) {    siz[u] = 1;    for( int t = head[u]; t; t = last[t] ) {        int v = dest[t];        if( v == fat[u] ) continue;        fat[v] = u;        dep[v] = dep[u] + 1;        val[v] = wght[t];        dfs1( v );        siz[u] += siz[v];        if( siz[v] > siz[son[u]] ) son[u] = v;    }}void dfs2( int u ) {    in[u] = ++idc;    rin[idc] = u;    if( son[u] ) {        top[son[u]] = top[u];        dfs2( son[u] );    }    for( int t = head[u]; t; t = last[t] ) {        int v = dest[t];        if( v == fat[u] || v == son[u] ) continue;        top[v] = v;        dfs2( v );    }}int query( int u, int v ) {    Info uca(0,0), cav(0,0);        //  u to lca, lca to v    while( top[u] != top[v] ) {        if( dep[top[u]] > dep[top[v]] ) {            uca = merge( uca, query_ba( root, 1, idc, in[top[u]], in[u] ) );            u = fat[top[u]];        } else {            cav = merge( query_ab( root, 1, idc, in[top[v]], in[v] ), cav );            v = fat[top[v]];        }    }    if( dep[u] > dep[v] ) {        uca = merge( uca, query_ba( root, 1, idc, in[v]+1, in[u] ) );    } else if( dep[v] > dep[u] ) {        cav = merge( query_ab( root, 1, idc, in[u]+1, in[v] ), cav );    }    Info ans = merge( uca, cav );    return ans.v;}int main() {    freopen("seqmod.in","r",stdin);    freopen("seqmod.out","w",stdout);    scanf( "%d%d", &n, &m );    for( int i = 1; i < n; i++ ) {        int u, v, d;        scanf( "%d%d%d", &u, &v, &d );        adde( u, v, d );        adde( v, u, d );    }    init();    fat[1] = 1;    dep[1] = 0;    dfs1( 1 );    top[1] = 1;    dfs2( 1 );    root = build( 1, idc );    while( m-- ) {        int u, v;        scanf( "%d%d", &u, &v );        printf( "%d\n", query(u,v) );    }}

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