SP1716 GSS3 - Can you answer these queries III

description

solution

设 $f_i:$ 前 $i$ 个数的最大子段和， $g_i:$ 以 $i$ 结尾的最大子段和

$g_i=\max(g_{i-1}+a_i,a_i)$

$f_i=max(f_{i-1},g_i)$

$max⁡\max$ 对于 $+$ 具有分配率max(a+c,b+c)=c+max(a,b)

所以广义的矩阵加速就是在原始的矩阵上 $∗→+*\rightarrow +$ ， $+→max⁡+\rightarrow \max$

为了套用矩阵乘法，改写一下 $g_i=\max(g_{i-1}+a_i,a_i,-inf),f_i=\max(f_{i-1},g_i,-inf)$
$[fi−1gi−10]×[0aiai−infaiai−inf−inf0]=[figi0]\begin{bmatrix} f_{i-1}\\ g_{i-1}\\ 0\\ \end{bmatrix} \times \begin{bmatrix} 0&a_i&a_i\\ -inf&a_i&a_i\\ -inf&-inf&0\\ \end{bmatrix}= \begin{bmatrix} f_i\\ g_i\\ 0 \end{bmatrix}$

code

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define maxn 50005
#define inf 0x3f3f3f3f
struct matrix {int c[3][3];matrix operator * ( const matrix &t ) const {matrix ans;for( int i = 0;i < 3;i ++ )for( int j = 0;j < 3;j ++ )ans.c[i][j] = -inf;for( int i = 0;i < 3;i ++ )for( int j = 0;j < 3;j ++ )for( int k = 0;k < 3;k ++ )ans.c[i][j] = max( ans.c[i][j], c[i][k] + t.c[k][j] );return ans;}
}t[maxn << 2], ret;
int n, Q;
int a[maxn];void New( int num, int pos ) {t[num].c[0][0] = t[num].c[0][1] = t[num].c[2][0] = t[num].c[2][1] = a[pos];t[num].c[0][2] = t[num].c[1][0] = t[num].c[1][2] = -inf;t[num].c[1][1] = t[num].c[2][2] = 0;
}void build( int num, int l, int r ) {if( l == r ) {New( num, l );return;}int mid = ( l + r ) >> 1;build( num << 1, l, mid );build( num << 1 | 1, mid + 1, r );t[num] = t[num << 1] * t[num << 1 | 1];
}void modify( int num, int l, int r, int pos ) {if( l == r ) {New( num, pos );return;}int mid = ( l + r ) >> 1;if( pos <= mid ) modify( num << 1, l, mid, pos );else modify( num << 1 | 1, mid + 1, r, pos );t[num] = t[num << 1] * t[num << 1 | 1];
}matrix query( int num, int l, int r, int L, int R ) {if( L <= l && r <= R ) return t[num];int mid = ( l + r ) >> 1;if( R <= mid ) return query( num << 1, l, mid, L, R );else if( mid < L ) return query( num << 1 | 1, mid + 1, r, L, R );else return query( num << 1, l, mid, L, R ) * query( num << 1 | 1, mid + 1, r, L, R );
}int main() {scanf( "%d", &n );for( int i = 1;i <= n;i ++ )scanf( "%d", &a[i] );build( 1, 1, n );scanf( "%d", &Q );while( Q -- ) {int opt, l, r;scanf( "%d %d %d", &opt, &l, &r );if( ! opt ) a[l] = r, modify( 1, 1, n, l );else {ret = query( 1, 1, n, l, r );printf( "%d\n", ret.c[2][1] );}}	return 0;
}

[NOIP2018 提高组] 保卫王国

description

solution

考虑对于一条链的简单转移，设 $dp_{i,0/1}:i$ 点不选/选的最小花费

矩阵形式：
$[fi−1,0fi−1,1]×[−inf0aiai]=[fi,0fi,1]\begin{bmatrix} f_{i-1,0}\\ f_{i-1,1}\\ \end{bmatrix} \times \begin{bmatrix} -inf&0\\ a_i&a_i\\ \end{bmatrix}= \begin{bmatrix} f_{i,0}\\ f_{i,1}\\ \end{bmatrix}$
在树上，就是普通的树形 $D P$ ，设 $dp_{i,0/1}:i$ 子树内 $i$ 点不选/选的最小花费

$dpi,0=∑j∈sonidpj,1dp_{i,0}=\sum_{j\in son_i}dp_{j,1}$

$dpi,1=∑j∈sonimin⁡(dpj,0,dpj,1)dp_{i,1}=\sum_{j\in son_i}\min(dp_{j,0},dp_{j,1})$

对于某个点转移是固定的且 $min⁡\min$ 对 $+$ 同样具有分配率

考虑动态 $D P$ ，树链剖分，轻重链断树，将矩阵放到线段树上合并加速

$gi,0=∑j∈soni,j≠MaxSonifj,1g_{i,0}=\sum_{j\in son_i,j≠MaxSon_i}f_{j,1}$

$gi,1=∑j∈soni,j≠MaxSonimin⁡(fj,0,fj,1)g_{i,1}=\sum_{j\in son_i,j≠MaxSon_i}\min(f_{j,0},f_{j,1})$

矩阵形式：
$[fMaxSoni,0fMaxSoni,1]×[−infgi,0gi,1gi,1]=[fi,0fi,1]\begin{bmatrix} f_{MaxSon_i,0}\\ f_{MaxSon_i,1}\\ \end{bmatrix} \times \begin{bmatrix} -inf&g_{i,0}\\ g_{i,1}&g_{i,1}\\ \end{bmatrix}= \begin{bmatrix} f_{i,0}\\ f_{i,1}\\ \end{bmatrix}$
一条重链顶点的 $D P$ 值就是这条链上所有矩阵“乘积”

~~具体细节详情请见完美代码，非常好懂，只是长了点而已~~
在这里插入图片描述

code

#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
using namespace std;
#define inf 1e18 
#define int long long
#define maxn 100005
vector < int > G[maxn];
char type[4];
int n, m, cnt;
int p[maxn], siz[maxn], fa[maxn], son[maxn], dfn[maxn], id[maxn], top[maxn], rnk[maxn], Left[maxn], Right[maxn], dep[maxn];
int f[maxn][2], g[maxn][2], ans[2][2], MS[2][2];
int matrix[maxn << 2][2][2];void dfs1( int u, int father ) {siz[u] = 1, fa[u] = father, dep[u] = dep[father] + 1, f[u][1] = p[u];for( auto v : G[u] ) {if( v == father ) continue;else dfs1( v, u );siz[u] += siz[v];f[u][0] += f[v][1];f[u][1] += min( f[v][0], f[v][1] );if( ! son[u] || siz[v] > siz[son[u]] )son[u] = v;}
}
/*
f[x][0/1]:x不选/选
f[x][0]+=f[son][1]
f[x][1]+=min(f[son][0],f[son][1])
g[x][0]:定义等同于f[x][0]但只含轻儿子
g[x][1]:定义等同于f[x][1]但只含轻儿子 
left[x]-right[x]:一条重链所管辖的区间(dfn序) 
*/
void dfs2( int u, int t ) {dfn[u] = Left[u] = Right[u] = ++ cnt, id[cnt] = u, top[u] = t;g[u][1] = p[u];if( ! son[u] ) return;else dfs2( son[u], t );Right[u] = Right[son[u]];for( auto v : G[u] ) {if( v == fa[u] || v == son[u] ) continue;else dfs2( v, v );g[u][0] += f[v][1];g[u][1] += min( f[v][0], f[v][1] );}
}
/*
动态dp
* -> +
+ -> min
*/
void pushup( int num ) {for( int i = 0;i < 2;i ++ )for( int j = 0;j < 2;j ++ )matrix[num][i][j] = inf;for( int i = 0;i < 2;i ++ )for( int j = 0;j < 2;j ++ )for( int k = 0;k < 2;k ++ )matrix[num][i][j] = min( matrix[num][i][j], matrix[num << 1][i][k] + matrix[num << 1 | 1][k][j] );
}
/*
|inf  A[u]|
|B[u] B[u]|
*
|f[maxson][0]|
|f[maxson][1]|
=
f[u][0]
f[u][1] 
*/
void build( int num, int l, int r ) {if( l == r ) {rnk[id[l]] = num;matrix[num][0][0] = inf;matrix[num][0][1] = g[id[l]][0];matrix[num][1][0] = matrix[num][1][1] = g[id[l]][1];return;}int mid = ( l + r ) >> 1;build( num << 1, l, mid );build( num << 1 | 1, mid + 1, r );pushup( num );
}void modify( int num, int l, int r, int pos ) {if( l == r ) return;int mid = ( l + r ) >> 1;if( pos <= mid ) modify( num << 1, l, mid, pos );else modify( num << 1 | 1, mid + 1, r, pos );pushup( num );
}void query( int num, int l, int r, int L, int R ) {if( L <= l && r <= R ) {for( int i = 0;i < 2;i ++ )for( int j = 0;j < 2;j ++ )MS[i][j] = inf;for( int i = 0;i < 2;i ++ )for( int j = 0;j < 2;j ++ )for( int k = 0;k < 2;k ++ )MS[i][j] = min( MS[i][j], ans[i][k] + matrix[num][k][j] );memcpy( ans, MS, sizeof( MS ) );return;}int mid = ( l + r ) >> 1;if( L <= mid ) query( num << 1, l, mid, L, R );if( mid < R ) query( num << 1 | 1, mid + 1, r, L, R );
}void modify( int x, int t ) {if( t == -1 ) {matrix[rnk[x]][0][0] = inf;matrix[rnk[x]][0][1] = g[x][0];matrix[rnk[x]][1][0] = matrix[rnk[x]][1][1] = g[x][1];}else if( t == 0 ) {//不能驻扎军队f[x][1]贡献必须无 设其转移矩阵inf 最后取min才会不选matrix[rnk[x]][0][0] = matrix[rnk[x]][1][0] = matrix[rnk[x]][1][1] = inf;matrix[rnk[x]][0][1] = g[x][0];}else {//必须驻扎军队f[x][1]贡献必须有 设f[x][0]转移矩阵inf 取min一定选f[x][1]matrix[rnk[x]][0][0] = matrix[rnk[x]][0][1] = inf;matrix[rnk[x]][1][0] = matrix[rnk[x]][1][1] = g[x][1];}modify( 1, 1, n, dfn[x] );//线段树的矩阵更新 while( x ) {//爬链x = top[x];ans[0][0] = ans[1][1] = 0;ans[0][1] = ans[1][0] = inf;//广义矩阵下的单位矩阵定义query( 1, 1, n, Left[x], Right[x] );//查询整条重链´ int IS = f[x][0], SI = f[x][1];f[x][0] = min( ans[0][0], ans[0][1] );f[x][1] = min( ans[1][0], ans[1][1] );if( x == 1 ) return;int y = fa[x];//轻链跳到重链上g[y][0] += ( f[x][1] - SI );//更新y(不选)的轻儿子贡献 减去原来的加上现在的f[x][1]g[y][1] += ( min( f[x][0], f[x][1] ) - min( SI, IS ) );//更新y(选)的轻儿子贡献减去原来的min加上现在的min(f[x][1],f[x][0])matrix[rnk[y]][0][1] = g[y][0];matrix[rnk[y]][1][0] = matrix[rnk[y]][1][1] = g[y][1];modify( 1, 1, n, dfn[y] );x = y;}
}signed main() {freopen( "defense.in", "r", stdin );freopen( "defense.out", "w", stdout );scanf( "%lld %lld %s", &n, &m, type );for( int i = 1;i <= n;i ++ )scanf( "%lld", &p[i] );	for( int i = 1, u, v;i < n;i ++ ) {scanf( "%lld %lld", &u, &v );G[u].push_back( v );G[v].push_back( u );}dfs1( 1, 0 );dfs2( 1, 1 );build( 1, 1, n );while( m -- ) {int a, x, b, y;scanf( "%lld %lld %lld %lld", &a, &x, &b, &y );if( dep[a] < dep[b] ) swap( a, b ), swap( x, y );
//先修改上面的点的话 后面在下面的点的修改其实会影响上面的点 那么之前上面的点的转移就是虚假的modify( a, x ), modify( b, y );if( min( f[1][0], f[1][1] ) >= inf ) printf( "-1\n" );else printf( "%lld\n", min( f[1][0], f[1][1] ) );modify( a, -1 ), modify( b, -1 );//询问独立最后要重置矩阵}return 0;
}