LuoguP4719 模板动态 DP(树形DP，矩阵加速，LCT)

Posted 2020-11-21 bingoyes

(n imes m)的算法谁都会吧，注意到每次修改影响的仅是一部分的信息，因此可思考优化。
将每个节点对应一个矩阵(egin{bmatrix} g[v][0] & g[v][0] \\ g[v][1] & -infty end{bmatrix}) ，从而 (egin{bmatrix} g[v][0] & g[v][0] \\ g[v][1] & -infty end{bmatrix} imes egin{bmatrix} f[son[u]][0] \\ f[son[u]][1] end{bmatrix} = egin{bmatrix} f[u][0] \\ f[u][1] end{bmatrix})，(LCT)虚实相生，维护子树信息

#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstring>
#define R(a,b,c) for(register int a = (b); a <= (c); ++a)
#define nR(a,b,c) for(register int a = (b); a >= (c); --a)
#define Fill(a,b) memset(a,b,sizeof(a))
#define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))
#define QWQ
#ifdef QWQ
#define D_e(x) cout << (#x) << " : " << x << "
"
#define D_e_Line printf("
----------------
")
#define FileOpen() freopen("in.txt", "r", stdin)
#define FileSave() freopen("out.txt","w", stdout)
#define Pause() system("pause")
#define TIME() fprintf(stderr, "TIME : %.3lfms
", clock() / CLOCKS_PER_SEC)
#endif
struct FastIO {
    template<typename ATP> inline FastIO& operator >> (ATP &x) {
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
        while(c >= '0' && c <= '9') x =x * 10 + (c ^ '0'), c = getchar();
        x = f == 1 ? x : -x;
        return *this;
    }
} io;
using namespace std;
template<typename ATP> inline ATP Max(ATP x, ATP y) {
    return x > y ? x : y;
}
template<typename ATP> inline ATP Min(ATP x, ATP y) {
    return x < y ? x : y;
}

const int N = 1e6 + 7;

struct Matrix {
    int mat[2][2];
    Matrix() {
        Fill(mat, 0);
    }
    inline void New(const int &A, const int &B) {
        mat[0][0] = mat[0][1] = A;
        mat[1][0] = B, mat[1][1] = -0x3f3f3f3f;
        /*
        [ g_u0, g_u0
          g_u1, -inf ]
        */
    }
    inline int Max(){
        return ::Max(mat[0][0], mat[1][0]); 
    }
    inline Matrix operator * (const Matrix &b) const {
        Matrix c;
        R(i,0,1){
            R(j,0,1){
                c.mat[i][j] = ::Max(mat[i][0] + b.mat[0][j], mat[i][1] + b.mat[1][j]);
            }
        }
        return c;
    }
};

struct Edge {
    int nxt, pre;
} e[N << 1];
int head[N], cntEdge;
inline void add(int u, int v) {
    e[++cntEdge] = (Edge){ head[u], v}, head[u] = cntEdge;
}

struct nod {
    int ch[2], fa, f[2];
    Matrix x; 
} t[N];
#define ls t[u].ch[0]
#define rs t[u].ch[1]
inline int Ident(int u) {
    return t[t[u].fa].ch[1] == u;
}
inline bool Isroot(int u) {
    return t[t[u].fa].ch[0] != u && t[t[u].fa].ch[1] != u;
}

inline void Pushup(int u) {
    t[u].x.New(t[u].f[0], t[u].f[1]);
    if(ls) t[u].x = t[ls].x * t[u].x;
    if(rs) t[u].x = t[u].x * t[rs].x;
}

inline void Rotate(int x) {
    int y = t[x].fa, z = t[y].fa, k = Ident(x);
    t[x].fa = z; if(!Isroot(y)) t[z].ch[Ident(y)] = x;
    t[y].ch[k] = t[x].ch[k ^ 1], t[t[x].ch[k ^ 1]].fa = y;
    t[x].ch[k ^ 1] = y, t[y].fa = x;
    Pushup(y), Pushup(x);
}

inline void Splay(int x) {
    while(!Isroot(x)){
        int y = t[x].fa;
        if(!Isroot(y))
            Ident(x) == Ident(y) ? Rotate(y) : Rotate(x);
        Rotate(x);
    }
    Pushup(x);
}

inline void Access(int u) {
    for(register int v = 0; u; v = u, u = t[u].fa){
        Splay(u);
        // the past comes to be my power, the future just lies
        if(rs){
            t[u].f[0] += t[rs].x.Max(); 
            t[u].f[1] += t[rs].x.mat[0][0];
        }
        if(v){
            t[u].f[0] -= t[v].x.Max();
            t[u].f[1] -= t[v].x.mat[0][0];
        }
        rs = v;
        Pushup(u);
    }
}

int val[N];
inline void DFS(int u, int father) {
    // a simple DP on tree, for the first blood
    t[u].f[1] = val[u];
    for(register int i = head[u]; i; i = e[i].nxt){
        int v = e[i].pre;
        if(v == father) continue;
        t[v].fa = u;
        DFS(v, u);
        t[u].f[0] += Max(t[v].f[0], t[v].f[1]);
        t[u].f[1] += t[v].f[0];
    }
    t[u].x.New(t[u].f[0], t[u].f[1]); // so the legend of mine is built
}

int main() {
//FileOpen();
//FileSave();
    int n, m;
    io >> n >> m;
    R(i,1,n){
        io >> val[i];
    }
    R(i,2,n){
        int u, v;
        io >> u >> v;
        add(u, v);
        add(v, u);
    }
    DFS(1, 0); // 1, our king
    while(m--){
        int x, newVal;
        io >> x >> newVal;
        Access(x); // you are my family now, every one should know you
        Splay(x); // be my king
        t[x].f[1] += newVal - val[x]; // the new up, the old down
        val[x] = newVal; // and so the new be the old
        Pushup(x); // you should use out your power, for out family
        Splay(1); // but when your power is out, when the old king returns, you are just a simple man
        printf("%d
", t[1].x.Max()); // as only him is the man who can tell the maximum value of us
    }
    
    return 0;
}