文章目錄

• Robot
• Product Sum
• Building Bridges
• Jump mission

Robot

BZOJ3938

機器人每次一旦改變速度，直到下一次改變速度為止

這一時間段內機器人的位置下標可以用一次函數表示

如果知道時刻$t_1$即將改變速度的機器人位置，以及最近的下一次機器人速度再次變化的時刻$t_2$

利用數學工具，可以算出這條線段的解析式（兩點式計算）

$(t_1,d)?(t_2,d+v(t_2?t_1))$

機器人的多次更換速度，那么一個機器人的位置其實就是若干條折線

將操作按照時間排序，記錄上一次同一個機器人的操作時間，速度等然后計算出線段

李超數就可以查詢$t$時刻的最遠值了（位置有正負所以計算后要取絕對值哦）

還有時間$t$的離散化，線段樹內存不允許開$[0,1e9]$這么大

#include <cstdio> #include <iostream> #include <algorithm> using namespace std; #define int long long #define inf 1e9 #define maxn 600005 #define lson num << 1 #define rson num << 1 | 1 struct node{int k, b;node(){}node( int K, int B ) { k = K, b = B; } }t[maxn << 2];struct query { int ti, k, x, opt; }q[maxn];int d[maxn], Time[maxn], V[maxn], lst_t[maxn], lst_v[maxn], T[maxn], val[maxn]; bool vis[maxn]; int m;void build( int num, int l, int r ) {t[num] = node( 0, 0 );if( l == r ) return;int mid = l + r >> 1;build( lson, l, mid );build( rson, mid + 1, r ); }int Fabs( int x ) { return x < 0 ? -x : x; }int calc( node l, int x ) { return Fabs( l.k * T[x] + l.b ); }bool cover( node lst, node now, int x ) { return calc( lst, x ) < calc( now, x ); }void insert( int num, int l, int r, node now ) {if( cover( t[num], now, l ) and cover( t[num], now, r ) ) {t[num] = now;return;}if( l == r ) return;int mid = l + r >> 1;if( cover( t[num], now, mid ) ) swap( t[num], now );if( cover( t[num], now, l ) ) insert( lson, l, mid, now );if( cover( t[num], now, r ) ) insert( rson, mid + 1, r, now ); }void modify( int num, int l, int r, int L, int R, node now ) {if( R < l or r < L ) return;if( L <= l and r <= R ) { insert( num, l, r, now ); return; }int mid = l + r >> 1;modify( lson, l, mid, L, R, now );modify( rson, mid + 1, r, L, R, now ); }int query( int num, int l, int r, int x ) {if( l == r ) return calc( t[num], x );int mid = l + r >> 1, ans;if( x <= mid ) ans = query( lson, l, mid, x );else ans = query( rson, mid + 1, r, x );return max( ans, calc( t[num], x ) ); }int find( int x ) { return lower_bound( T + 1, T + m + 1, x ) - T; }signed main() {int n, Q; char opt[10];scanf( "%lld %lld", &n, &Q ); build( 1, 0, maxn );for( int i = 1;i <= n;i ++ ) {scanf( "%lld", &d[i] );modify( 1, 0, maxn, 0, 0, node( 0, d[i] ) );}T[++ m] = 0;for( int i = 1;i <= Q;i ++ ) {scanf( "%lld %s", &q[i].ti, opt );T[++ m] = q[i].ti;if( opt[0] == 'c' ) {q[i].opt = 0;scanf( "%lld %lld", &q[i].k, &q[i].x );lst_t[i] = Time[q[i].k], Time[q[i].k] = q[i].ti;lst_v[i] = V[q[i].k], V[q[i].k] = q[i].x;}else q[i].opt = 1;}sort( T + 1, T + m + 1 );m = unique( T + 1, T + m + 1 ) - T - 1;for( int i = 1;i <= Q;i ++ )if( q[i].opt ) continue;else {int dt = q[i].ti - lst_t[i];int y = dt * lst_v[i] + d[q[i].k];int k = lst_v[i];int b = y - q[i].ti * k;modify( 1, 0, m, find( lst_t[i] ), find( q[i].ti ), node( k, b ) );d[q[i].k] = y;}for( int i = Q;i;i -- )if( vis[q[i].k] ) continue;else {vis[q[i].k] = 1;int dt = inf - q[i].ti;int y = dt * q[i].x + d[q[i].k];int k = q[i].x;int b = y - inf * k;modify( 1, 0, m, find( q[i].ti ), m, node( k, b ) );}for( int i = 1;i <= n;i ++ )if( ! vis[i] ) modify( 1, 0, m, 0, m, node( 0, d[i] ) );for( int i = 1;i <= Q;i ++ )if( q[i].opt ) printf( "%lld\n", query( 1, 0, m, find( q[i].ti ) ) );return 0; }

Product Sum

CF631E

令$S={\sum }_{i=1}^{n}i?a_i$

考慮當位置$i$的數插到位置$j$的時，權值的變化

• $j<i$，位置$i$的數往前插

  • $?k\in [1,j)?(i,n]k?a_k$沒有變化

  • $?k\in [j,i)$ 每個數都會往后移一位，$k?a_k\leftarrow (k+1)?a_k$，$S$增加${\sum }_{k=j}^{i?1}{a}_k$

  • $i$前移$i?j$位到$j$，$i?a_i\leftarrow j?a_i$，$S$減少$(i?j)a_i$

  • 設$f_i$ ：考慮第$i$個位置往前移動到某個位置時的最大權值，$su{m}_i={\sum }_{j=1}^i{a}_j$

  • 從小到大枚舉i

  • $$f_i=\min \{S+su{m}_{i?1}?su{m}_{j?1}+(i?j)a_i\}=S+su{m}_{i?1}+i?a_i+\min \{?j?a_i?su{m}_{j?1}\}$$

  • $?j$看做直線斜率$k$，$?su{m}_{j?1}$看做直線截距$b$，$a_i$就是因變量$x$

• $j>i$，位置$i$的數往后插

  • $?k\in [1,i)?(j,n]k?a_k$沒有變化

  • $?k\in (i,j]$ 每個數都會往前移一位，$k?a_k\leftarrow (k?1)?a_k$，$S$減少${\sum }_{k=i+1}^j{a}_k$

  • $i$后移$j?i$位到$j$，$i?a_i\leftarrow j?a_i$，$S$增加$(j?i)a_i$

  • 設$f_i$ ：考慮第$i$個位置往移動后到某個位置時的最大權值，$su{m}_i={\sum }_{j=1}^i{a}_j$

  • 從大到小枚舉i

  • $$f_i=\min \{S?(su{m}_j?su{m}_i)+(j?i)a_i\}=S+su{m}_i?i?a_i+\min \{j?a_i?su{m}_j\}$$

  • $j$看做直線斜率$k$，$?su{m}_j$看做直線截距$b$，$a_i$就是因變量$x$

#include <cstdio> #include <iostream> using namespace std; #define int long long #define inf 1e18 #define maxn 1000000 #define lson num << 1 #define rson num << 1 | 1 struct node {int k, b;node(){}node( int K, int B ) { k = K, b = B; } }t[( maxn + 5 ) << 3];void build( int num, int l, int r ) {t[num] = node( 0, -inf );if( l == r ) return;int mid = l + r >> 1;build( lson, l, mid );build( rson, mid + 1, r ); }int calc( node l, int x ) { return l.k * x + l.b; }int cover( node lst, node now, int x ) { return calc( lst, x ) < calc( now, x ); }void insert( int num, int l, int r, node now ) {if( cover( t[num], now, l ) and cover( t[num], now, r ) ) {t[num] = now;return;}if( l == r ) return;int mid = l + r >> 1;if( cover( t[num], now, mid ) ) swap( t[num], now );if( cover( t[num], now, l ) ) insert( lson, l, mid, now );if( cover( t[num], now, r ) ) insert( rson, mid + 1, r, now ); }int query( int num, int l, int r, int x ) {if( l == r ) return calc( t[num], x );int mid = l + r >> 1, ans = 0;if( x <= mid ) ans = query( lson, l, mid, x );else ans = query( rson, mid + 1, r, x );return max( ans, calc( t[num], x ) ); }int A[maxn], sum[maxn], ans, ret; int n; signed main() {scanf( "%lld", &n );for( int i = 1;i <= n;i ++ ) scanf( "%lld", &A[i] );for( int i = 1;i <= n;i ++ ) sum[i] = sum[i - 1] + A[i];for( int i = 1;i <= n;i ++ ) ans += i * A[i];ret = ans;build( 1, -maxn, maxn );insert( 1, 1, n, node( 1, 0 ) );for( int i = 2;i <= n;i ++ ) {ret = max( ret, ans + sum[i - 1] - A[i] * i + query( 1, -maxn, maxn, A[i] ) );insert( 1, -maxn, maxn, node( i, - sum[i - 1] ) );}build( 1, -maxn, maxn );insert( 1, -maxn, maxn, node( n, -sum[n] ) );for( int i = n - 1;i;i -- ) {ret = max( ret, ans + sum[i] - A[i] * i + query( 1, -maxn, maxn, A[i] ) );insert( 1, -maxn, maxn, node( i, - sum[i] ) );}printf( "%lld\n", ret );return 0; }

Building Bridges

LOJ2483

設$d{p}_i:$ 使$1?i$聯通的最小花費

考慮暴力$dp$轉移，枚舉橋建筑在$j$和$i$根柱子之間

$d{p}_i=\min \{d{p}_j+{\sum }_{k=j+1}^{i?1}{w}_k+(h_i?h_j{)}^2\}$

前綴和優化，$su{m}_i={\sum }_{j=1}^i{w}_j$
$$\begin{gathered} d{p}_i=\min \{d{p}_j+su{m}_{i?1}?su{m}_j+h_i^2?2h_i{h}_j+h_j^2\} \\ =su{m}_{i?1}+h_i^2+\min \{?2h_j?h_i+d{p}_j?su{m}_j+h_j^2\} \end{gathered}$$

#include <cstdio> #include <iostream> using namespace std; #define int long long #define maxn 1000000 #define lson num << 1 #define rson num << 1 | 1 #define inf 1e18 struct node {int k, b;node(){}node( int K, int B ) { k = K, b = B; } }t[( maxn + 5 ) << 2];int calc( node l, int x ) { return l.k * x + l.b; }bool cover( node lst, node now, int x ) { return calc( lst, x ) > calc( now, x ); }void build( int num, int l, int r ) {t[num] = node( 0, inf );if( l == r ) return;int mid = l + r >> 1;build( lson, l, mid );build( rson, mid + 1, r ); }void insert( int num, int l, int r, node now ) {if( cover( t[num], now, l ) and cover( t[num], now, r ) ) {t[num] = now;return;}if( l == r ) return;int mid = l + r >> 1;if( cover( t[num], now, mid ) ) swap( t[num], now );if( cover( t[num], now, l ) ) insert( lson, l, mid, now );if( cover( t[num], now, r ) ) insert( rson, mid + 1, r, now ); }int query( int num, int l, int r, int x ) {if( l == r ) return calc( t[num], x );int mid = l + r >> 1, ans;if( x <= mid ) ans = query( lson, l, mid, x );else ans = query( rson, mid + 1, r, x );return min( ans, calc( t[num], x ) ); }int h[maxn], w[maxn], dp[maxn], sum[maxn];signed main() {int n;scanf( "%lld", &n );for( int i = 1;i <= n;i ++ ) scanf( "%lld", &h[i] );for( int i = 1;i <= n;i ++ ) scanf( "%lld", &w[i] );for( int i = 1;i <= n;i ++ ) sum[i] = sum[i - 1] + w[i];build( 1, 0, maxn );insert( 1, 0, maxn, node( -2 * h[1], -sum[1] + h[1] * h[1] ) );for( int i = 2;i <= n;i ++ ) {dp[i] = sum[i - 1] + h[i] * h[i] + query( 1, 0, maxn, h[i] );insert( 1, 0, maxn, node( -2 * h[i], dp[i] - sum[i] + h[i] * h[i] ) );}printf( "%lld\n", dp[n] );return 0; }

Jump mission

CodeChef

設$d{p}_i:$ 在$i$座山停留的最小花費

考慮暴力$dp$轉移，枚舉上一次停留在$j$山
$$d{p}_i=\min \{d{p}_j+A_i+h_i^2?2h_i{h}_j+h_j^2\}=A_i+h_i^2+\min \{?2h_j?h_i+d{p}_j+h_j^2\}$$
從小到大枚舉$i$，利用前面枚舉過的山當$j$，這只能保證$j<i$的條件

然而本題多了一個$P_j<P_i$的限制

這個時候只有獻出神祭——樹套樹，將$P$利用樹狀數組嵌套在李超線段樹外面

自然內存就會多一些，看著開就行了

#include <cstdio> #include <iostream> using namespace std; #define int long long #define inf 5e18 #define maxn 600000 struct node {int k, b;node(){ k = 0, b = inf; }node( int K, int B ) { k = K, b = B; } }t[maxn * 80]; int lson[maxn * 80], rson[maxn * 80]; int cnt;int calc( node l, int x ) { return l.k * x + l.b; }bool cover( node lst, node now, int x ) { return calc( lst, x ) > calc( now, x ); }void insert( int &num, int l, int r, node now ) {if( ! num ) num = ++ cnt, t[num] = node( 0, inf );if( cover( t[num], now, l ) and cover( t[num], now, r ) ) {t[num] = now;return;}if( l == r ) return;int mid = ( l + r ) >> 1;if( cover( t[num], now, mid ) ) swap( t[num], now );if( cover( t[num], now, l ) ) insert( lson[num], l, mid, now );if( cover( t[num], now, r ) ) insert( rson[num], mid + 1, r, now ); }int query( int num, int l, int r, int x ) {if( ! num ) return inf;if( l == r ) return calc( t[num], x );int mid = ( l + r ) >> 1, ans;if( x <= mid ) ans = query( lson[num], l, mid, x );else ans = query( rson[num], mid + 1, r, x );return min( ans, calc( t[num], x ) ); }int p[maxn], a[maxn], h[maxn], dp[maxn], root[maxn]; int n;int lowbit( int i ) { return i & -i; }void add( int i, node l ) {for( ;i <= n;i += lowbit( i ) ) insert( root[i], 1, maxn, l ); }int query( int i, int x ) {int ans = inf;for( ;i;i -= lowbit( i ) ) ans = min( ans, query( root[i], 1, maxn, x ) );return ans; }signed main() {scanf( "%lld", &n );for( int i = 1;i <= n;i ++ ) scanf( "%lld", &p[i] );for( int i = 1;i <= n;i ++ ) scanf( "%lld", &a[i] );for( int i = 1;i <= n;i ++ ) scanf( "%lld", &h[i] );dp[1] = a[1], add( p[1], node( -2 * h[1], h[1] * h[1] + dp[1] ) );for( int i = 2;i <= n;i ++ ) {dp[i] = a[i] + h[i] * h[i] + query( p[i], h[i] );add( p[i], node( - 2 * h[i], h[i] * h[i] + dp[i] ) );}printf( "%lld\n", dp[n] );return 0; }

李超數注意兩點即可

• 推出能夠轉化成直線解析式的轉移方程式
• 確定線段樹點的區間范圍，究竟是哪個值做線段樹區間，是否需要離散化

總結

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