HDU 5293 Tree chain problem

树状数组 + dp

设$f_i$表示以$i$为根的子树中的能选取的最大和,$sum_x$表示$\sum_{f_y}$  ($y$是$x$的一个儿子),这样子我们把所有给出的链按照两点的$lca$分组,对于每一个点$x$,$sum_x$显然是一个$f_x$的一个备选答案,而当有树链的$lca$正好是$x$时,我们发现$sum_x + w + \sum_{sum_t} - \sum_{f_t}$($w$代表这条树链能产生的价值,$t$是树链上的一个点)。

那么我们只要能快速计算出这两个$\sum$就可以转移了,其实用两个树状数组维护$dfs$序即可。

具体可以参照代码。

时间复杂度$O(Tnlogn)$。

Code:

#include 
#include 
#include 
using namespace std;

const int N = 1e5 + 5;
const int Lg = 20;

int testCase, n, m, f[N], sum[N], dfsc, in[N], out[N];
int tot, head[N], fa[N][Lg], dep[N];
vector <int> vec[N];

struct Edge {
    int to, nxt;
} e[N << 1];

inline void add(int from, int to) {
    e[++tot].to = to;
    e[tot].nxt = head[from];
    head[from] = tot;
}

struct Item {
    int u, v, lca, val;
} a[N];

inline void read(int &X) {
    X = 0; char ch = 0; int op = 1;
    for(; ch > '9' || ch < '0'; ch = getchar())
        if(ch == '-') op = -1;
    for(; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

struct Bit {
    int s[N << 1];

    #define lowbit(p) (p & (-p))

    inline void clear() {
        memset(s, 0, sizeof(s));
    }

    inline void modify(int p, int val) {
        for(; p <= 2 * n; p += lowbit(p))
            s[p] += val;
    }

    inline int query(int p) {
        int res = 0;
        for(; p > 0; p -= lowbit(p))
            res += s[p];
        return res;
    }

} bit1, bit2;

inline void swap(int &x, int &y) {
    int t = x; x = y; y = t;
}

inline void chkMax(int &x, int y) {
    if(y > x) x = y;
}

void dfs(int x, int fat, int depth) {
    fa[x][0] = fat, dep[x] = depth;
    in[x] = ++dfsc;
    for(int i = 1; i <= 18; i++)
        fa[x][i] = fa[fa[x][i - 1]][i - 1];

    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(y == fat) continue;
        dfs(y, x, depth + 1);
    }
    out[x] = ++dfsc;
}

inline int getLca(int x, int y) {
    if(dep[x] < dep[y]) swap(x, y); 
    for(int i = 18; i >= 0; i--)
        if(dep[fa[x][i]] >= dep[y])
            x = fa[x][i];
    if(x == y) return x;
    for(int i = 18; i >= 0; i--)
        if(fa[x][i] != fa[y][i])
            x = fa[x][i], y = fa[y][i];
    return fa[x][0];
}

void solve(int x, int fat) {
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(y == fat) continue;
        solve(y, x);

        sum[x] += f[y];
    }

    chkMax(f[x], sum[x]);
    for(unsigned int i = 0; i < vec[x].size(); i++) {
        int id = vec[x][i];
        int now = bit1.query(in[a[id].v]) + bit1.query(in[a[id].u]) - bit2.query(in[a[id].u]) - bit2.query(in[a[id].v]) + sum[x];
        chkMax(f[x], now + a[id].val);
    }

    bit1.modify(in[x], sum[x]), bit1.modify(out[x], -sum[x]);
    bit2.modify(in[x], f[x]), bit2.modify(out[x], -f[x]);
}

int main() {
    for(read(testCase); testCase--; ) {
        tot = 0; memset(head, 0, sizeof(head));
        read(n), read(m);
        for(int x, y, i = 1; i < n; i++) {
            read(x), read(y);
            add(x, y), add(y, x);
        }

        dfsc = 0; dfs(1, 0, 1);

        for(int i = 1; i <= n; i++) vec[i].clear();
        for(int i = 1; i <= m; i++) {
            read(a[i].u), read(a[i].v), read(a[i].val);
            a[i].lca = getLca(a[i].u, a[i].v);
            vec[a[i].lca].push_back(i);
        }

        memset(f, 0, sizeof(f));
        memset(sum, 0, sizeof(sum));
        bit1.clear(), bit2.clear();
        solve(1, 0);

        printf("%d\n", f[1]);
    }

    return 0;
}
View Code

 

转载于:https://www.cnblogs.com/CzxingcHen/p/9819619.html

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