[Luogu P6021]洪水（ddp）

题目描述

小 A 走到一个山脚下，准备给自己造一个小屋。这时候，小 A 的朋友（op，又叫管理员）打开了创造模式，然后飞到山顶放了格水。于是小 A 面前出现了一个瀑布。作为平民的小 A 只好老实巴交地爬山堵水。那么问题来了：我们把这个瀑布看成是一个

$设$dp_i$代表把$i$与它的子树内的叶节点断开所需的最少距离，有$dp_u=min(sum dp_v, a_u)$ 还是很套路地把它拆成和重儿子相关和不相关。$

$$g_u=sum_{v e son(u)} dp_v,dp_u=min(g_u+dp_{son(u)},a_u+0)$。$

Code

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

typedef long long ll;

const int N = 200010;

namespace IO{
    template <typename T> void read(T &x) {
        T f = 1;
        char ch = getchar();
        for (; !isdigit(ch); ch = getchar()) if (ch == '-') f = -1;
        for (x = 0; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
        x *= f;
    }
    template <typename T> void write(T x) {
        if (x > 9) write(x / 10);
        putchar(x % 10 + '0');
    }
    template <typename T> void print(T x) {
        if (x < 0) x = -x, putchar('-');
        write(x);
        putchar('
');
    }
} using namespace IO;

int n, m;
ll a[N];

namespace QXX{
    struct node{
        int pre, to;
    }edge[N << 1];
    int head[N], tot;
    inline void add(int u, int v) {
        edge[++tot] = node{head[u], v};
        head[u] = tot;
    }
}using namespace QXX;

namespace MATRIX{
    template <typename T> void cmin(T &x, T y) {if (y < x) x = y;}
    template <typename T> void cmax(T &x, T y) {if (y > x) x = y;}
    struct Matrix{
        ll arr[3][3];
    };
    inline void init(Matrix &X) {
        memset(X.arr, 0x3f, sizeof(X.arr));
    }
    inline Matrix Mul(Matrix X, Matrix Y) {
        Matrix Z;
        init(Z);
        for (int i = 1; i <= 2; i++)
            for (int j = 1; j <= 2; j++)
                for (int k = 1; k <= 2; k++)
                    cmin(Z.arr[i][j], X.arr[i][k] + Y.arr[k][j]);
        return Z;
    }
    inline void deburg(Matrix X) {
        for (int i = 1; i <= 2; i++) {
            for (int j = 1; j <= 2; j++) {
                cout << X.arr[i][j] << " ";
            } cout << endl;
        } cout << endl;
    }
}using namespace MATRIX;

namespace TCP{
    Matrix A[N];
    ll dp[N];
    int dfn[N], pos[N], top[N], End[N], num;
    int sz[N], fa[N], son[N];
    void dfs1(int x) {
        sz[x] = 1;
        for (int i = head[x]; i; i = edge[i].pre) {
            if (edge[i].to == fa[x]) continue;
            fa[edge[i].to] = x;
            dfs1(edge[i].to);
            sz[x] += sz[edge[i].to];
            if (sz[edge[i].to] > sz[son[x]]) son[x] = edge[i].to;
        }
    }
    void dfs2(int x, int chain) {
        dfn[x] = ++num, top[x] = chain, pos[dfn[x]] = x;
        init(A[x]);
        A[x].arr[1][1] = 0, A[x].arr[1][2] = a[x];
        A[x].arr[2][2] = 0;
        if (son[x])
            dfs2(son[x], chain);
        else
            End[chain] = dfn[x], A[x].arr[1][1] = a[x], A[x].arr[2][1] = 0;
        for (int i = head[x]; i; i = edge[i].pre) {
            if (edge[i].to == fa[x] || edge[i].to == son[x]) continue;
            dfs2(edge[i].to, edge[i].to);
            A[x].arr[1][1] += dp[edge[i].to];
        }
        if (son[x]) dp[x] = min(a[x], A[x].arr[1][1] + dp[son[x]]);
        else dp[x] = a[x];
    }
}using namespace TCP;

namespace Segment_Tree{
    struct Segment{
        Matrix val;
    }tr[N << 2];
    inline void push_up(int p) {tr[p].val = Mul(tr[p << 1].val, tr[p << 1 | 1].val);}
    void build(int p, int l, int r) {
        if (l == r) {
            tr[p].val = A[pos[l]];
            return;
        }
        int mid = (l + r) >> 1;
        build(p << 1, l, mid);
        build(p << 1 | 1, mid + 1, r);
        push_up(p);
    }
    void change(int p, int l, int r, int Pos) {
        if (l == r) {
            tr[p].val = A[pos[l]];
            return;
        }
        int mid = (l + r) >> 1;
        if (Pos <= mid) change(p << 1, l, mid, Pos);
        else change(p << 1 | 1, mid + 1, r, Pos);
        push_up(p);
    }
    Matrix query(int p, int l, int r, int L, int R) {
        if (L <= l && r <= R) {
            return tr[p].val;
        }
        int mid = (l + r) >> 1;
        if (R <= mid) return query(p << 1, l, mid, L, R);
        else if (L > mid) return query(p << 1 | 1, mid + 1, r, L, R);
        else return Mul(query(p << 1, l, mid, L, mid), query(p << 1 | 1, mid + 1, r, mid + 1, R));
    }
}using namespace Segment_Tree;

inline void update(int x, ll val) {//将节点 a 的权值修改为 b 
    A[x].arr[1][2] += (val - a[x]);
    if (!son[x]) A[x].arr[1][1] = A[x].arr[1][2];
    a[x] = val;
    while (x) {
        Matrix bef = query(1, 1, n, dfn[top[x]], End[top[x]]);
        change(1, 1, n, dfn[x]);
        Matrix aft = query(1, 1, n, dfn[top[x]], End[top[x]]);
        x = fa[top[x]];
        A[x].arr[1][1] += aft.arr[1][1] - bef.arr[1][1];
        
    }
}

int main() {
    read(n);
    for (int i = 1; i <= n; i++) read(a[i]);
    for (int i = 1, u, v; i < n; i++) read(u), read(v), add(u, v), add(v, u);
    read(m);
    dfs1(1);
    dfs2(1, 1);
    build(1, 1, n);
    while (m--) {
        char opt[2];
        int x;
        scanf("%s", opt + 1); read(x);
        if (opt[1] ==  'Q') {
            Matrix ans = query(1, 1, n, dfn[x], End[top[x]]);
            print(ans.arr[1][1]);
        } else {
            int val;
            read(val);
            update(x, a[x] + val);
        }
    }
    return 0;
}

[Luogu P6021]洪水（ddp）

题目描述

Solution

Code