POJ3237 Tree 树链剖分 边权

POJ3237 Tree 树链剖分 边权

传送门：http://poj.org/problem?id=3237

题意：

n个点的，n-1条边

修改单边边权

将a->b的边权取反

查询a->b边权最大值

题解：

修改边权就查询点的深度大的点，用大的点去存这条边的边权，其余的就和点权的是一样的了

取反操作用线段树维护，区间最大值取反就是区间最小值，区间最小值取反就是区间最大值

所以维护两颗线段树即可，lazy标记表示覆盖单边的边权

代码：

#include <set>
#include <map>
#include <cmath>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const int maxn = 1e5 + 5;
const int INF = 0x3f3f3f3f;
struct EDGE {
    int v, nxt, w;
} edge[maxn << 1];
int head[maxn], tot;
void add_edge(int u, int v, int w) {
    edge[tot].v = v;
    edge[tot].w = w;
    edge[tot].nxt = head[u];
    head[u] = tot++;
}
int sz[maxn], dep[maxn], son[maxn], id[maxn], Rank[maxn], cnt, fa[maxn], top[maxn];
int d[maxn];
void dfs1(int u, int f, int cnt) {
    fa[u] = f;
    dep[u] = cnt;
    sz[u] = 1;
    son[u] = 0;
    int tmp = 0;
    for(int i = head[u]; i != -1; i = edge[i].nxt) {
        int v = edge[i].v;
        if(v != f) {
            dfs1(v, u, cnt + 1);
            if(tmp < sz[v]) {
                son[u] = v;
                tmp = sz[v];
            }
            sz[u] += sz[v];
        }

    }
}
void dfs2(int u, int tp) {
    top[u] = tp;
    id[u] = ++cnt;
    Rank[cnt] = u;
    if(son[u]) dfs2(son[u], tp);
    for(int i = head[u]; i != -1; i = edge[i].nxt) {
        int v = edge[i].v;
        if(v == fa[u]) continue;
        if(v == son[u]) {
            d[id[v]] = edge[i].w;
            continue;
        }
        dfs2(v, v);
        d[id[v]] = edge[i].w;
    }
}
void prebuild() {
    dfs1(1, 0, 0);
    dfs2(1, 1);
}

int Max[maxn << 2];
int Min[maxn << 2];
// int sum[maxn<<2];
int lazy[maxn];

void push_up(int rt) {
    Max[rt] = max(Max[ls], Max[rs]);
    Min[rt] = min(Min[ls], Min[rs]);

}
void build(int l, int r, int rt) {
    lazy[rt] = 1;
    if(l == r) {
        Max[rt] = Min[rt] = d[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
    push_up(rt);
}
void push_down(int rt) {
    if(lazy[rt] == -1) {
        lazy[ls] = -lazy[ls];
        lazy[rs] = -lazy[rs];
        lazy[rt] = 1;
        swap(Max[ls], Min[ls]);
        Max[ls] *= -1;
        Min[ls] *= -1;
        swap(Max[rs], Min[rs]);
        Max[rs] *= -1;
        Min[rs] *= -1;
    }
}
void update_pos(int pos, int val, int l, int r, int rt) {
    if(l == r) {
        lazy[rt] = 1;
        Max[rt] = Min[rt] = val;
        return;
    }
    push_down(rt);
    int mid = (l + r) >> 1;
    if(pos <= mid) update_pos(pos, val, lson);
    else update_pos(pos, val, rson);
    push_up(rt);
}
void update(int L, int R, int l, int r, int rt) {
    if(L <= l && r <= R) {
        lazy[rt] = -lazy[rt];
        swap(Max[rt], Min[rt]);
        Max[rt] *= -1;
        Min[rt] *= -1;
        return;
    }
    push_down(rt);
    int mid = (l + r) >> 1;
    if(L <= mid) update(L, R, lson);
    if(R > mid) update(L, R, rson);
    push_up(rt);
}
int query(int L, int R, int l, int r, int rt) {
    if(L <= l && r <= R) {
        return Max[rt];
    }
    push_down(rt);
    int mid = (l + r) >> 1;
    int ans = -INF;
    if(L <= mid) ans = max(ans, query(L, R, lson));
    if(R > mid) ans = max(ans, query(L, R, rson));
    return ans;
}
void change(int u, int v) {
    while(top[u] != top[v]) {
        if(dep[top[u]] < dep[top[v]]) {
            swap(u, v);
        }
        update(id[top[u]], id[u], 1, cnt, 1);
        u = fa[top[u]];
    }
    if(u != v) {
        if(dep[u] > dep[v]) swap(u, v);
        update(id[son[u]], id[v], 1, cnt, 1);
    }
}
void Query(int u, int v) {
    int ans = -INF;
    while(top[u] != top[v]) {
        if(dep[top[u]] < dep[top[v]]) {
            swap(u, v);
        }
        ans = max(ans, query(id[top[u]], id[u], 1, cnt, 1));
        u = fa[top[u]];
    }
    if(u != v) {
        if(dep[u] > dep[v]) swap(u, v);
        ans = max(ans, query(id[son[u]], id[v], 1, cnt, 1));
    }
    printf("%d\n", ans);
}
int u[maxn], v[maxn], c[maxn];


int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    int n, T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        memset(head, -1, sizeof(head));
        tot = cnt = 0;
        for(int i = 1; i < n; i++) {
            scanf("%d%d%d", &u[i], &v[i], &c[i]); //要用数组保存
            add_edge(u[i], v[i], c[i]);
            add_edge(v[i], u[i], c[i]);
        }
        prebuild();
        build(1, cnt, 1);
        char op[20];
        int a, b;
        while(1) {
            scanf("%s", op);
            if(op[0] == 'D') break;
            scanf("%d%d", &a, &b);
            if(op[0] == 'C') {
                int tmp = dep[u[a]] > dep[v[a]] ? u[a] : v[a]; //找出深度大的那个点
                update_pos(id[tmp], b, 1, cnt, 1); //更新进入深度大的点那条边
            } else if(op[0] == 'N') change(a, b);
            else if(op[0] == 'Q') Query(a, b);
        }
    }
    return 0;

}