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spoj 375 QTREE - Query on a tree 树链剖分

时间:2015-12-29 00:55:28      阅读:197      评论:0      收藏:0      [点我收藏+]

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题目链接

给一棵树, 每条边有权值, 两种操作, 一种是将一条边的权值改变, 一种是询问u到v路径上最大的边的权值。

树链剖分模板。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 10005;
int head[maxn*2], son[maxn], sz[maxn], deep[maxn], top[maxn], w[maxn], f[maxn], cnt, num;
int edge[maxn][3], maxx[maxn<<2];
struct node
{
    int to, nextt;
}e[maxn*2];
void init() {
    mem1(head);
    num = cnt = 0;
}
void add(int u, int v, int w) {
    e[num].to = v, e[num].nextt = head[u], head[u] = num++;
}
void dfs1(int u, int fa) {
    sz[u] = 1;
    deep[u] = deep[fa]+1;
    son[u] = -1;
    f[u] = fa;
    for(int i = head[u]; ~i; i = e[i].nextt) {
        int v = e[i].to;
        if(v == fa)
            continue;
        dfs1(v, u);
        sz[u] += sz[v];
        if(son[u]==-1||sz[v]>sz[son[u]])
            son[u] = v;
    }
}
void dfs2(int u, int tp) {
    w[u] = ++cnt, top[u] = tp;
    if(~son[u])
        dfs2(son[u], tp);
    for(int i = head[u]; ~i; i = e[i].nextt) {
        int v = e[i].to;
        if(v == f[u]||v == son[u])
            continue;
        dfs2(v, v);
    }
}
void pushUp(int rt) {
    maxx[rt] = max(maxx[rt<<1], maxx[rt<<1|1]);
}
void update(int p, int val, int l, int r, int rt) {
    if(l == r) {
        maxx[rt] = val;
        return ;
    }
    int m = l+r>>1;
    if(p<=m)
        update(p, val, lson);
    else
        update(p, val, rson);
    pushUp(rt);
}
int query(int L, int R, int l, int r, int rt) {
    if(L<=l&&R>=r) {
        return maxx[rt];
    }
    int m = l+r>>1, ret = 0;
    if(L<=m)
        ret = max(ret, query(L, R, lson));
    if(R>m)
        ret = max(ret, query(L, R, rson));
    return ret;
}
int find(int u, int v) {
    int f1 = top[u], f2 = top[v], ret = 0;
    while(f1 != f2) {
        if(deep[f1]<deep[f2]) {
            swap(f1, f2);
            swap(u, v);
        }
        ret = max(ret, query(w[f1], w[u], 1, cnt, 1));
        u = f[f1];
        f1 = top[u];
    }
    if(u == v)
        return ret;
    if(deep[u]>deep[v])
        swap(u, v);
    return max(ret, query(w[son[u]], w[v], 1, cnt, 1));
}
int main()
{
    int t, n, u, v, val;
    cin>>t;
    while(t--) {
        scanf("%d", &n);
        init();
        deep[0] = 0;
        for(int i = 1; i<n; i++) {
            scanf("%d%d%d", &u, &v, &val);
            add(u, v, val);
            add(v, u, val);
            edge[i][0] = u, edge[i][1] = v, edge[i][2] = val;
        }
        dfs1(1, 0);
        dfs2(1, 1);
        for(int i = 1; i<n; i++) {
            if(deep[edge[i][0]]>deep[edge[i][1]]) {
                swap(edge[i][0], edge[i][1]);
            }
            update(w[edge[i][1]], edge[i][2], 1, cnt, 1);
        }
        char s[10];
        while(scanf("%s", s)) {
            if(s[0] == D)
                break;
            int x, y;
            scanf("%d%d", &x, &y);
            if(s[0]==Q) {
                printf("%d\n", find(x, y));
            } else {
                update(w[edge[x][1]], y, 1, cnt, 1);
            }
        }
    }
    return 0;
}

 

spoj 375 QTREE - Query on a tree 树链剖分

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原文地址:http://www.cnblogs.com/yohaha/p/5084462.html

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