POJ-2763 Housewife Wind

Housewife Wind

树链剖分

利用树中每个点都只有一个父节点的性质，将边权化为点权，然后树链剖分 LCA，套一个线段树维护权值

注意在同一链的时候，LCA 的点权是不计算在内的

这题卡时间卡的很紧，如果 TLE，尝试改成 scanf printf

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 1e5 + 10;
#define pii pair<int, int>
int hson[maxn], siz[maxn], fa[maxn], dep[maxn];
int dfn[maxn], top[maxn], rnk[maxn];
int w[maxn], id[maxn], p[maxn];
int tr[maxn << 2];
vector<pii>gra[maxn];

void build(int now, int l, int r)
{
    if(l == r)
    {
        tr[now] = w[p[rnk[l]]];
        return;
    }
    int mid = l + r >> 1;
    build(now << 1, l, mid);
    build(now << 1 | 1, mid + 1, r);
    tr[now] = tr[now << 1] + tr[now << 1 | 1];
}

void update(int now, int l, int r, int x, int val)
{
    if(l == r)
    {
        tr[now] = val;
        return;
    }
    int mid = l + r >> 1;
    if(x <= mid)
        update(now << 1, l, mid, x, val);
    else
        update(now << 1 | 1, mid + 1, r, x, val);
    tr[now] = tr[now << 1] + tr[now << 1 | 1]; 
}

int query(int now, int l, int r, int L, int R)
{
    if(L <= l && r <= R)
        return tr[now];
    int mid = l + r >> 1;
    int ans = 0;
    if(L <= mid)
        ans += query(now << 1, l, mid, L, R);
    if(R > mid)
        ans += query(now << 1 | 1, mid + 1, r, L, R);
    return ans;
}

void dfs1(int now, int cnt)
{
    siz[now] = 1;
    hson[now] = 0;
    dep[now] = cnt;
    for(int i=0; i<gra[now].size(); i++)
    {
        int nex = gra[now][i].first;
        int x = gra[now][i].second;
        if(nex == fa[now]) continue;
        fa[nex] = now;
        id[x] = nex;
        p[nex] = x;
        dfs1(nex, cnt + 1);
        siz[now] += siz[nex];
        if(siz[hson[now]] < siz[nex])
            hson[now] = nex;
    }
}

int tp = 0;
void dfs2(int now, int t)
{
    tp++;
    dfn[now] = tp;
    rnk[tp] = now;
    top[now] = t;
    if(hson[now])
    {
        dfs2(hson[now], t);
        for(int i=0; i<gra[now].size(); i++)
        {
            int nex = gra[now][i].first;
            if(nex == fa[now] || nex == hson[now]) continue;
            dfs2(nex, nex);
        }
    }
}

void init(int n, int rt)
{
    tp = 0;
    fa[rt] = rt;
    dfs1(rt, 1);
    dfs2(rt, rt);
    build(1, 1, n);
}

int LCA(int a, int b, int n)
{
    int ans = 0;
    while(top[a] != top[b])
    {
        if(dep[top[a]] < dep[top[b]]) swap(a, b);
        ans += query(1, 1, n, dfn[top[a]], dfn[a]);
        a = fa[top[a]];
    }
    if(dfn[a] > dfn[b]) swap(a, b);
    if(a != b)
        ans += query(1, 1, n, dfn[a] + 1, dfn[b]);
    return ans;
}

int main()
{
    int n, q, rt;
    scanf("%d%d%d", &n, &q, &rt);
    for(int i=1; i<n; i++)
    {
        int x, y;
        scanf("%d%d%d", &x, &y, &w[i]);
        gra[x].push_back(make_pair(y, i));
        gra[y].push_back(make_pair(x, i));
    }
    init(n, rt);
    while(q--)
    {
        int t;
        scanf("%d", &t);
        if(t == 1)
        {
            int i, x;
            scanf("%d%d", &i, &x);
            update(1, 1, n, dfn[id[i]], x);
        }
        else
        {
            int u;
            scanf("%d", &u);
            printf("%d\n", LCA(rt, u, n));
            swap(rt, u);
        }
    }
    return 0;
}

标签：now,2763,int,dfn,maxn,ans,POJ,include,Wind
来源： https://www.cnblogs.com/dgsvygd/p/16450977.html