每日练习2021.9.22

题目：

1. ZOJ 3261

2. HDU 4725

3. BZOJ 4052

1. ZOJ 3261 Connections in Galaxy War

思路很简单，将询问离线下来反向操作，将删边转化为加边，就可以使用并查集简单处理了。本题和洛谷P1197几乎是相同的。 代码: 

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 10;
int fa[ maxn ], rnk[ maxn ];
int find( int x ) {
  if ( x == fa[ x ] ) return x;
  return fa[ x ] = find( fa[ x ] );
}
void merge( int x, int y ) {
  x = find( x ), y = find( y );
  if ( x == y ) return;
  if ( rnk[ x ] < rnk[ y ] )
    fa[ x ] = y;
  else if ( rnk[ x ] > rnk[ y ] )
    fa[ y ] = x;
  else
    fa[ max( x, y ) ] = min( x, y );
}
int n, m, t;
struct Q {
  int op, x, y;
} q[ maxn * 5 ];
int ans[ maxn * 5 ];
struct edge {
  int u, v, hs;
} e[ maxn * 2 ];
unordered_map< int, int > bk;

int main( ) {
  int T = 0;
  while ( cin >> n ) {
    if ( T++ ) cout << '\n';
    bk.clear( );
    for ( int i = 0; i < n; i++ ) {
      fa[ i ] = i;
      cin >> rnk[ i ];
    }
    cin >> m;
    for ( int i = 1; i <= m; i++ ) {
      int u, v;
      cin >> u >> v;
      if ( u > v ) swap( u, v );
      e[ i ] = edge { u, v, 100000 * u + v };
    }
    cin >> t;
    for ( int i = 1; i <= t; i++ ) {
      string str;
      cin >> str;
      if ( str[ 0 ] == 'q' ) {
        q[ i ].op = q[ i ].y = 0;
        cin >> q[ i ].x;
      } else {
        q[ i ].op = 1;
        int u, v;
        cin >> u >> v;
        if ( u > v ) swap( u, v );
        q[ i ] = Q { 1, u, v };
        bk[ 100000 * u + v ] = 1;
      }
    }
    for ( int i = 1; i <= m; i++ ) {
      int u = e[ i ].u, v = e[ i ].v;
      if ( ! bk[ e[ i ].hs ] ) merge( u, v );
    }
    int tot = 0;
    for ( int i = t; i >= 1; i-- ) {
      if ( q[ i ].op == 0 ) {
        if ( rnk[ find( q[ i ].x ) ] <= rnk[ q[ i ].x ] )
          ans[ ++tot ] = -1;
        else
          ans[ ++tot ] = find( q[ i ].x );
      } else {
        merge( q[ i ].x, q[ i ].y );
      }
    }
    for ( int i = tot; i >= 1; i-- )
      cout << ans[ i ] << '\n';
  }
  return 0;
}

2. HDU 4725 The Shortest Path in Nya Graph

在每层添加两个超级源点，分别作为每层的源点和汇点，边权为0，然后跑dijkstra，如果只添加一个，那么每层所有点之间的最短路就变成了0，假了

代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 10;

struct node {
  int to, w;
};
vector< node > G[ maxn ];
void add( int u, int v, int w ) {
  G[ u ].push_back( node { v, w } );
}
struct hnode {
  int d, u;
  bool operator<( const hnode &x ) const {
    return d > x.d;
  }
};
int d[ maxn ], vis[ maxn ];
int n, m, c;


void dijkstra( ) {
  memset( d, 0x3f, sizeof( d ) );
  memset( vis, 0, sizeof( vis ) );
  d[ 1 ] = 0;
  priority_queue< hnode > q;
  q.push( hnode { 0, 1 } );
  while ( ! q.empty( ) ) {
    int u = q.top( ).u;
    q.pop( );
    if ( vis[ u ] ) continue;
    vis[ u ] = 1;
    for ( int i = 0; i < G[ u ].size( ); i++ ) {
      node e = G[ u ][ i ];
      if ( d[ e.to ] > d[ u ] + e.w ) {
        d[ e.to ] = d[ u ] + e.w;
        q.push( hnode { d[ e.to ], e.to } );
      }
    }
  }
}

int main( ) {
  ios::sync_with_stdio( false );
  cin.tie( nullptr ), cout.tie( nullptr );
  int T, kse = 1;
  cin >> T;
  while ( T-- ) {
    cin >> n >> m >> c;
    for ( int i = 1; i < maxn; i++ )
      G[ i ].clear( );
    for ( int i = 1; i <= n; i++ ) {
      int x;
      cin >> x;
      add( x + n, i, 0 ), add( i, x + n * 2, 0 );
    }
    for ( int i = 1; i <= m; i++ ) {
      int u, v, w;
      cin >> u >> v >> w;
      add( u, v, w ), add( v, u, w );
    }
    for ( int i = 1; i < n; i++ ) {
      add( i + n * 2, i + 1 + n, c );
      add( i + 1 + n * 2, i + n, c );
    }
    dijkstra( );
    cout << "Case #" << kse++ << ": ";
    if ( d[ n ] == 0x3f3f3f3f )
      cout << "-1\n";
    else
      cout << d[ n ] << '\n';
  }
  return 0;
}

3. BZOJ 4052 Magical GCD

和HDU 5726差不多，随手写了个st表+二分，数据比较水，除了老毛子的OJ都没被卡掉，先就这样吧 >__<

#include <bits/stdc++.h>
using namespace std;
constexpr int maxn = 1e5 + 10;
typedef long long ll;
ll st[ maxn ][ 20 ];
int Log[ maxn ];
ll n;
inline ll read( ) {
  ll x = 0;
  char ch = getchar( );
  while ( ch < '0' || ch > '9' )
    ch = getchar( );
  while ( ch >= '0' && ch <= '9' )
    x = x * 10 + ch - 48, ch = getchar( );
  return x;
}
ll query( int l, int r ) {
  int k = Log[ r - l + 1 ];
  return __gcd( st[ l ][ k ], st[ r - ( 1 << k ) + 1 ][ k ] );
}
int main( ) {
  Log[ 1 ] = 0, Log[ 2 ] = 1;
  for ( int i = 3; i < maxn; i++ )
    Log[ i ] = Log[ i / 2 ] + 1;
  ll _ = read( );
  while ( _-- ) {
    n = read( );
    for ( int i = 1; i <= n; i++ ) {
      st[ i ][ 0 ] = read( );
    }

    for ( int j = 1; j <= 17; j++ ) {
      for ( int i = 1; i + ( 1 << j ) - 1 <= n; i++ )
        st[ i ][ j ] =
            __gcd( st[ i ][ j - 1 ], st[ i + ( 1 << j - 1 ) ][ j - 1 ] );
    }

    ll ans = -1;
    for ( int i = 1; i <= n; i++ ) {
      int j = i;
      while ( j <= n ) {
        ll x = query( i, j );
        int l = j, r = n + 1;
        while ( r - l > 1 ) {
          int mid = ( r + l ) >> 1;
          if ( query( i, mid ) == x )
            l = mid;
          else
            r = mid;
        }
        ans = max( ans, ( l - i + 1 ) * query( i, l ) );
        j = l + 1;
      }
    }
    printf( "%lld\n", ans );
  }
  return 0;
}