动态DP
动态DP
最大权独立集
修改树上的点权, 并查询最大权独立集, 即选出一些不相邻的点, 使他们点权和最大
点数和操作数\(n, m(\le 10^5)\)
假如不修改的话:
- 设\(f[x][0/1]\)表示这个点选不选时, 整棵子树的最大点权和
- \(f[u][0]=\sum \max(f[v][0/1])\)
- \(f[u][1]=\sum f[v][0]+Val(u)\)
那么要修改查询怎么办呢?
- 熟练剖分
- 套路是利用矩阵表示\(DP\)方程, 然后利用矩阵的结合律和线段树的区间维护, 来快速修改查询区间的值
- 设\(f[x]\)为这个点的真答案, \(g[x]\)为这个点除了重子树外的假答案
- 这样每次修改就只需要更新到根的路径中,轻儿子连到的那个重链的点的\(g\)值即可
以这道题具体来说:
设\(g[u][0/1]\)表示这个点除了重子树外取与不取的最大权值和
将矩阵\(C[i][j]=\sum A[i][k]*B[k][j]\)改为\(C[i][j]=\max(A[i][k]+B[k][j])\)
至于为什么魔改之后仍满足结合律,参考floyd的代码实现就可以了
- 三个
for枚举并取\(\max\),显然两次矩乘先后乘都没关系
继续:
\[\begin{bmatrix}
g[u][0] & g[u][0] \\
g[u][1] & -\infty \\
\end{bmatrix}
*
\begin{bmatrix}
f[hson][0] \\
f[hson][1]
\end{bmatrix}
=
\begin{bmatrix}
f[u][0] \\
f[u][1]
\end{bmatrix}
\]
那么在一条重链上:
\[\prod_{i=1}^{n-1}
\begin{bmatrix}
g[i][0] & g[i][0] \\
g[i][1] & -\infty \\
\end{bmatrix}
*
\begin{bmatrix}
f[n][0] \\
f[n][1]
\end{bmatrix}
=
\begin{bmatrix}
f[1][0] \\
f[1][1]
\end{bmatrix}
\]
由于在记录一个\(f\)矩阵太烦了, 思考怎么用\(g\)表示\(f\)
由于\(g[n][0/1]=f[n][0/1]\)
\[\begin{bmatrix}
g[n][0] & g[n][0] \\
g[n][1] & -\infty \\
\end{bmatrix}
*
\begin{bmatrix}
0 \\
-\infty \\
\end{bmatrix}
=
\begin{bmatrix}
f[n][0] \\
f[n][1] \\
\end{bmatrix}
\]
于是变成
\[\prod_{i=1}^{n}
\begin{bmatrix}
g[i][0] & g[i][0] \\
g[i][1] & -\infty \\
\end{bmatrix}
*
\begin{bmatrix}
0 \\
-\infty
\end{bmatrix}
=
\begin{bmatrix}
f[1][0] \\
f[1][1]
\end{bmatrix}
\]
那么显然可以用线段树维护左边那个矩阵连乘, 支持单点修改, 根的值查询即可
查询完答案后乘上那个转移用的矩阵就完事了
总结一下:
对于修改一个点\(x\)的权值, 他只影响他到根的路径上的点:
- 根据\(g\)函数的性质, 重新修改它
- 修改这个点到这条重链顶端的区间
- 记录修改前后的链顶的值, 用于更新链顶父亲的\(g\)
- 跳到链顶的父亲, 循环直到根
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
typedef long long LL;
const int MAXN = 1e5 + 10;
const LL INF = 1e14;
inline LL in()
{
LL x = 0, flag = 1; char ch = getchar();
while (ch < '0' || ch > '9') { if (ch == '-') flag = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
return x * flag;
}
int n, m;
LL a[MAXN], f[MAXN][2], g[MAXN][2];
int nume, head[MAXN];
struct Adj { int nex, to; } adj[MAXN << 1];
void addedge(int from, int to)
{
adj[++ nume] = (Adj) { head[from], to };
head[from] = nume;
}
struct Matrix
{
LL val[2][2];
Matrix () { memset(val, 0, sizeof val); }
Matrix (LL a, LL b, LL c, LL d) { val[0][0] = a, val[0][1] = b, val[1][0] = c, val[1][1] = d;}
Matrix (int x) { val[0][0] = val[0][1] = g[x][0], val[1][0] = g[x][1], val[1][1] = -INF; }
Matrix operator * (const Matrix B)
{
Matrix C;
for (int i = 0; i < 2; ++ i)
for (int j = 0; j < 2; ++ j)
for (int k = 0; k < 2; ++ k)
C.val[i][j] = max(C.val[i][j], val[i][k] + B.val[k][j]);
return C;
}
void print() { printf("%lld %lld\n%lld %lld\n", val[0][0], val[0][1], val[1][0], val[1][1]); }
} I;
int fa[MAXN], siz[MAXN], hson[MAXN];
void DFS1(int u) // fa, siz, hson
{
siz[u] = 1;
for (int i = head[u]; i; i = adj[i].nex)
{
int v = adj[i].to;
if (v == fa[u]) continue;
fa[v] = u;
DFS1(v);
siz[u] += siz[v];
if (siz[v] > siz[hson[u]]) hson[u] = v;
}
}
int ind, id[MAXN], aid[MAXN], top[MAXN], btm[MAXN];
void DFS2(int u, int t) // id, aid, top, btm
{
aid[id[u] = ++ ind] = u; top[u] = t;
if (hson[u]) DFS2(hson[u], t), btm[u] = btm[hson[u]];
else btm[u] = u;
for (int i = head[u]; i; i = adj[i].nex)
{
int v = adj[i].to;
if (v == fa[u] || v == hson[u]) continue;
DFS2(v, v);
}
}
#define ls (now << 1)
#define rs (now << 1 | 1)
#define mid ((l + r) >> 1)
Matrix t[MAXN << 2];
void build(int now, int l, int r)
{
if (l == r) return (void) (t[now] = Matrix(aid[l]));
build(ls, l, mid); build(rs, mid + 1, r);
t[now] = t[ls] * t[rs];
}
void modify(int now, int l, int r, int x)
{
if (x < l || r < x) return;
if (l == r) return (void) (t[now] = Matrix(aid[x]));
modify(ls, l, mid, x); modify(rs, mid + 1, r, x);
t[now] = t[ls] * t[rs];
}
Matrix query(int now, int l, int r, int x, int y)
{
if (x <= l && r <= y) return t[now];
if (y <= mid) return query(ls, l, mid, x, y);
else if (x >= mid + 1) return query(rs, mid + 1, r, x, y);
else return query(ls, l, mid, x, y) * query(rs, mid + 1, r, x, y);
}
#undef ls
#undef rs
#undef mid
void init(int u) // f[0] = sum(max(f[v0], f[v1])), f[1] = a[u] + sum(f[v0])
{
g[u][1] = f[u][1] = a[u];
for (int i = head[u]; i; i = adj[i].nex)
{
int v = adj[i].to;
if (v == fa[u]) continue;
init(v);
f[u][0] += max(f[v][0], f[v][1]);
f[u][1] += f[v][0];
if (v != hson[u])
{
g[u][0] += max(f[v][0], f[v][1]);
g[u][1] += f[v][0];
}
}
}
void uppath(int x, LL y)
{
g[x][1] += y;
Matrix now; LL pre0, pre1;
while (x)
{
pre0 = f[top[x]][0], pre1 = f[top[x]][1];
modify(1, 1, n, id[x]);
now = query(1, 1, n, id[top[x]], id[btm[x]]) * I;
f[top[x]][0] = now.val[0][0];
f[top[x]][1] = now.val[1][0];
x = fa[top[x]];
g[x][1] += now.val[0][0] - pre0;
g[x][0] += max(now.val[0][0], now.val[1][0]) - max(pre0, pre1);
}
}
int main()
{
I = Matrix(0, -INF, -INF, 0);
n = in(), m = in();
for (int i = 1; i <= n; ++ i) a[i] = in();
for (int i = 1; i < n; ++ i)
{
int x = in(), y = in();
addedge(x, y); addedge(y, x);
}
DFS1(1); DFS2(1, 1);
init(1); build(1, 1, n);
Matrix ret;
while (m --)
{
LL x = in(), y = in();
uppath(x, y - a[x]); a[x] = y;
ret = query(1, 1, n, id[top[1]], id[btm[1]]) * I;
printf("%lld\n", max(ret.val[0][0], ret.val[1][0]));
}
return 0;
}
/*201907191515~1809 */
NOIp2019 D2T3 保卫王国
同样
定义矩阵乘法\(C[i][j]=\max(A[i][k]+B[k][j])\)
\[\begin{bmatrix}
\infty & g[u][0] \\
g[u][1] & g[u][1] \\
\end{bmatrix}
*
\begin{bmatrix}
f[hson][0] \\
f[hson][1]
\end{bmatrix}
=
\begin{bmatrix}
f[u][0] \\
f[u][1]
\end{bmatrix}
\]
\[\begin{bmatrix}
\infty & g[u][0] \\
g[u][1] & g[u][1] \\
\end{bmatrix}
*
\begin{bmatrix}
0 & \infty\\
\infty & 0\\
\end{bmatrix}
=
\begin{bmatrix}
不关键 & f[n][0] \\
不关键 & f[n][1] \\
\end{bmatrix}
\]
\[ans=
\prod_{i=1}^{n}
\begin{bmatrix}
\infty & g[u][0] \\
g[u][1] & g[u][1] \\
\end{bmatrix}
*
\begin{bmatrix}
0 & \infty\\
\infty & 0\\
\end{bmatrix}
=
\begin{bmatrix}
不关键 & f[1][0] \\
不关键 & f[1][1] \\
\end{bmatrix}
\]
那么如果一个点必须选, 就给他的值减去\(\infty\),必须不选同理
假如两个点都必须不选,而算出来的答案\(>\infty\), 说明他们两个有连边, 所以算法只能至少选择其中一个, 输出\(-1\)即可
然后再修改回去
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
typedef long long LL;
const int MAXN = 1e5 + 10;
const LL INF = 1e14;
inline LL in()
{
LL x = 0, flag = 1; char ch = getchar();
while (ch < '0' || ch > '9') { if (ch == '-') flag = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
return x * flag;
}
int n, m;
LL a[MAXN], f[MAXN][2], g[MAXN][2];
int nume, head[MAXN];
struct Adj { int nex, to; } adj[MAXN << 1];
void addedge(int from, int to)
{
adj[++ nume] = (Adj) { head[from], to };
head[from] = nume;
}
struct Matrix
{
LL val[2][2];
Matrix () { memset(val, 0x3f3f3f, sizeof val); }
Matrix (LL a, LL b, LL c, LL d) { val[0][0] = a, val[0][1] = b, val[1][0] = c, val[1][1] = d;}
Matrix (int x) { val[0][0] = INF, val[0][1] = g[x][0], val[1][0] = val[1][1] = g[x][1]; }
Matrix operator * (const Matrix B)
{
Matrix C;
for (int i = 0; i < 2; ++ i)
for (int j = 0; j < 2; ++ j)
for (int k = 0; k < 2; ++ k)
C.val[i][j] = min(C.val[i][j], val[i][k] + B.val[k][j]);
return C;
}
void print() { printf("%lld %lld\n%lld %lld\n", val[0][0], val[0][1], val[1][0], val[1][1]); }
} I;
int fa[MAXN], siz[MAXN], hson[MAXN];
void DFS1(int u) // fa, siz, hson
{
siz[u] = 1;
for (int i = head[u]; i; i = adj[i].nex)
{
int v = adj[i].to;
if (v == fa[u]) continue;
fa[v] = u;
DFS1(v);
siz[u] += siz[v];
if (siz[v] > siz[hson[u]]) hson[u] = v;
}
}
int ind, id[MAXN], aid[MAXN], top[MAXN], btm[MAXN];
void DFS2(int u, int t) // id, aid, top, btm
{
aid[id[u] = ++ ind] = u; top[u] = t;
if (hson[u]) DFS2(hson[u], t), btm[u] = btm[hson[u]];
else btm[u] = u;
for (int i = head[u]; i; i = adj[i].nex)
{
int v = adj[i].to;
if (v == fa[u] || v == hson[u]) continue;
DFS2(v, v);
}
}
#define ls (now << 1)
#define rs (now << 1 | 1)
#define mid ((l + r) >> 1)
Matrix t[MAXN << 2];
void build(int now, int l, int r)
{
if (l == r) return (void) (t[now] = Matrix(aid[l]));
build(ls, l, mid); build(rs, mid + 1, r);
t[now] = t[ls] * t[rs];
}
void modify(int now, int l, int r, int x)
{
if (x < l || r < x) return;
if (l == r) return (void) (t[now] = Matrix(aid[x]));
modify(ls, l, mid, x); modify(rs, mid + 1, r, x);
t[now] = t[ls] * t[rs];
}
Matrix query(int now, int l, int r, int x, int y)
{
if (x <= l && r <= y) return t[now];
if (y <= mid) return query(ls, l, mid, x, y);
else if (x >= mid + 1) return query(rs, mid + 1, r, x, y);
else return query(ls, l, mid, x, y) * query(rs, mid + 1, r, x, y);
}
#undef ls
#undef rs
#undef mid
void init(int u) // f[1] = sum(min(f[v0], f[v1])) + a[u], f[0] = sum(f[v1])
{
g[u][1] = f[u][1] = a[u];
for (int i = head[u]; i; i = adj[i].nex)
{
int v = adj[i].to;
if (v == fa[u]) continue;
init(v);
f[u][1] += min(f[v][0], f[v][1]);
f[u][0] += f[v][1];
if (v != hson[u])
{
g[u][1] += min(f[v][0], f[v][1]);
g[u][0] += f[v][1];
}
}
}
void uppath(int x, LL y)
{
g[x][1] += y;
Matrix now; LL pre0, pre1;
while (x)
{
pre0 = f[top[x]][0], pre1 = f[top[x]][1];
modify(1, 1, n, id[x]);
now = query(1, 1, n, id[top[x]], id[btm[x]]) * I;
f[top[x]][0] = now.val[0][1];
f[top[x]][1] = now.val[1][1];
x = fa[top[x]];
g[x][0] += now.val[1][1] - pre1;
g[x][1] += min(now.val[0][1], now.val[1][1]) - min(pre0, pre1);
}
}
char fuck[5];
int main()
{
I = Matrix(0, INF, INF, 0);
n = in(), m = in(); scanf("%s", fuck);
for (int i = 1; i <= n; ++ i) a[i] = in();
for (int i = 1; i < n; ++ i)
{
int x = in(), y = in();
addedge(x, y); addedge(y, x);
}
DFS1(1); DFS2(1, 1);
init(1); build(1, 1, n);
Matrix ret;
while (m --)
{
LL x = in(), xx = in(), y = in(), yy = in();
if (xx == 0) uppath(x, INF); else uppath(x, -INF);
if (yy == 0) uppath(y, INF); else uppath(y, -INF);
ret = query(1, 1, n, id[top[1]], id[btm[1]]) * I;
if (xx == 0 && yy == 0 && min(ret.val[0][1], ret.val[1][1]) > INF) printf("-1\n");
else printf("%lld\n", min(ret.val[0][1], ret.val[1][1]) + (xx + yy) * INF);
if (xx == 0) uppath(x, -INF); else uppath(x, INF);
if (yy == 0) uppath(y, -INF); else uppath(y, INF);
}
return 0;
}
还可以直接套用最大全独立集, 不过容易写错.
