洛谷P4719 【模板】动态dp(ddp LCT)

题意

题目链接

Sol

动态dp板子题。有些细节还没搞懂，待我研究明白后再补题解。。。

#include<bits/stdc++.h>
#define LL long long 
using namespace std;
const int MAXN = 1e5 + 10, INF = INT_MAX;
template<typename A, typename B> inline bool chmax(A &x, B y) {
    return x < y ? x = y, 1 : 0;
}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M, a[MAXN], f[MAXN][2];
struct Ma {
    LL m[3][3];
    Ma() {
        memset(m, -0x3f, sizeof(m));
    }
    Ma operator * (const Ma &rhs) const {
        Ma ans;
        for(int i = 1; i <= 2; i++)
            for(int j = 1; j <= 2; j++)
                for(int k = 1; k <= 2; k++)
                    chmax(ans.m[i][j], m[i][k] + rhs.m[k][j]);
        return ans;
    }
}val[MAXN], sum[MAXN];
vector<int> v[MAXN];
int ch[MAXN][2], fa[MAXN];
#define ls(x) ch[x][0]
#define rs(x) ch[x][1]
void update(int k) {
    sum[k] = val[k];
    if(rs(k)) sum[k] = sum[k] * sum[rs(k)];
    if(ls(k)) sum[k] = sum[ls(k)] * sum[k];
}
bool ident(int x) {
    return rs(fa[x]) == x;
}
bool isroot(int x) {
    return ls(fa[x]) != x && rs(fa[x]) != x;
}
void connect(int x, int _fa, int tag) {
    fa[x] = _fa;
    ch[_fa][tag] = x;
}
void rotate(int x) {
    int y = fa[x], r = fa[y], yson = ident(x), rson = ident(y);
    int b = ch[x][yson ^ 1];
    fa[x] = r;
    if(!isroot(y)) connect(x, r, rson);
    connect(b, y, yson);
    connect(y, x, yson ^ 1);
    update(y); update(x);
}
void splay(int x) {
    for(int y = fa[x]; !isroot(x); rotate(x), y = fa[x])
        if(!isroot(y))
            rotate(ident(y) == ident(x) ? y : x);
}
void access(int x) {
    for(int y = 0; x; x = fa[y = x]) {
        splay(x);
        if(rs(x)) {
            val[x].m[1][1] += max(sum[rs(x)].m[1][1], sum[rs(x)].m[2][1]);
            val[x].m[2][1] += sum[rs(x)].m[1][1];
        }
        if(y) {
            val[x].m[1][1] -= max(sum[y].m[1][1], sum[y].m[2][1]);
            val[x].m[2][1] -= sum[y].m[1][1];
        }
        val[x].m[1][2] = val[x].m[1][1];
        rs(x) = y;
        update(x);
    }
}
int Modify(int p, int v) {
    access(p); splay(p);
    val[p].m[2][1] -= a[p] - v; 
    update(p);
    a[p] = v;
    splay(1);
    return max(sum[1].m[1][1], sum[1].m[2][1]);
}
void dfs(int x, int _fa) {
    fa[x] = _fa;
    f[x][1] = a[x];
    for(auto &to : v[x]) {
        if(to == _fa) continue;
        dfs(to, x);
        f[x][0] += max(f[to][0], f[to][1]);
        f[x][1] += f[to][0];
    }
    val[x].m[1][1] = f[x][0]; val[x].m[1][2] = f[x][0];
    val[x].m[2][1] = f[x][1]; val[x].m[2][2] = -INF;
    update(x);
//  sum[x] = val[x];
}
int main() {
    //freopen("a.in", "r", stdin);
    N = read(); M = read();
    for(int i = 1; i <= N; i++) a[i] = read();
    for(int i = 1; i < N; i++) {
        int x = read(), y = read();
        v[x].push_back(y);
        v[y].push_back(x);
    }
    dfs(1, 0);
    while(M--) {
        int x = read(), y = read();
        printf("%d\n", Modify(x, y));
    }
    return 0;
}

标签：LCT,洛谷,val,rs,int,sum,ddp,fa,void
来源： https://www.cnblogs.com/zwfymqz/p/10425674.html