树链剖分+线段树 poj2763 Housewife Wind

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题意：2种操作，一种路径上的值统一修改，一种是询问路径上数字连续区间个数。

思路：树链剖分+线段树合并。这道题目主要就是难在用树链剖分上套线段树合并后，因为整条链被分成了很多条短的，把这些短的也要按照顺序合并。

又因为其实是从左边和右边链的最底端向上执行的，所以应该把左右的分开，然后就是更接近根节点的点是区间的左区间，远离的点是右区间。

主要的问题就在于路径上的合并，我的代码这一部分写的比较搓。。

#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck(x) cout<<"["<<x<<"]"
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

const int MX = 1e5 + 5;
const int INF = 0x3f3f3f3f;

struct Edge {
    int v, nxt;
} E[MX << 2];
int Head[MX], h_r;
void edge_init() {
    h_r = 0;
    memset(Head, -1, sizeof(Head));
}
void edge_add(int u, int v) {
    E[h_r].v = v;
    E[h_r].nxt = Head[u];
    Head[u] = h_r++;
}


int S[MX << 2], col[MX << 2], SL[MX << 2], SR[MX << 2], A[MX], B[MX];
void push_up(int rt, int m) {
    S[rt] = S[rt << 1] + S[rt << 1 | 1];
    SL[rt] = SL[rt << 1]; SR[rt] = SR[rt << 1 | 1];
    if(SR[rt << 1] == SL[rt << 1 | 1]) S[rt]--;
}
void push_down(int rt) {
    if(col[rt] != -1) {
        col[rt << 1] = col[rt << 1 | 1] = col[rt];
        SL[rt << 1] = SR[rt << 1] = col[rt];
        SL[rt << 1 | 1] = SR[rt << 1 | 1] = col[rt];
        S[rt << 1] = S[rt << 1 | 1] = 1;
        col[rt] = -1;
    }
}
void build(int l, int r, int rt) {
    col[rt] = -1;
    if(l == r) {
        S[rt] = 1;
        SL[rt] = SR[rt] = A[l];
        return;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    push_up(rt, m);
}
void update(int L, int R, int x, int l, int r, int rt) {
    if(L <= l && r <= R) {
        S[rt] = 1;
        SL[rt] = SR[rt] = col[rt] = x;
        return;
    }
    push_down(rt);
    int m = (l + r) >> 1;
    if(L <= m) update(L, R, x, lson);
    if(R > m) update(L, R, x, rson);
    push_up(rt, m);
}
int query(int L, int R, int l, int r, int rt, int &la, int &ra) {
    if(L <= l && r <= R) {
        la = SL[rt]; ra = SR[rt];
        return S[rt];
    }
    push_down(rt);
    int m = (l + r) >> 1, ret = 0;
    int lla = -1, lra = -1, rla = -1, rra = -1;
    if(L <= m) ret += query(L, R, lson, lla, lra);
    if(R > m) ret += query(L, R, rson, rla, rra);
    if(L <= m && m < R && SR[rt << 1] == SL[rt << 1 | 1]) ret--;
    la = lla != -1 ? lla : rla;
    ra = rra != -1 ? rra : lra;
    return ret;
}


int fa[MX], top[MX], siz[MX], son[MX], dep[MX], id[MX], rear;
/*fa父节点，top重链开头起点，siz子树大小，son重儿子，dep深度，id新编号*/

/*第一次DFS找到重边，并维护好siz,son,fa,dep*/
void DFS1(int u, int f, int d) {
    fa[u] = f; dep[u] = d;
    son[u] = 0; siz[u] = 1;
    for(int i = Head[u]; ~i; i = E[i].nxt) {
        int v = E[i].v;
        if(v == f) continue;
        DFS1(v, u, d + 1);
        siz[u] += siz[v];
        if(siz[son[u]] < siz[v]) {
            son[u] = v;
        }
    }
}
/*将重边编号好，维护id*/
void DFS2(int u, int tp) {
    top[u] = tp;
    id[u] = ++rear;
    if(son[u]) DFS2(son[u], tp);
    for(int i = Head[u]; ~i; i = E[i].nxt) {
        int v = E[i].v;
        if(v == fa[u] || v == son[u]) continue;
        DFS2(v, v);
    }
}
/*用来给点编号，以及建立线段树，要用id编号建树*/
void HLD_presolve() {
    rear = 0;
    DFS1(1, 0, 1);
    DFS2(1, 1);
    for(int i = 1; i <= rear; i++) {
        A[id[i]] = B[i];
    }
    build(1, rear, 1);
}
/*修改，要注意使用id编号修改*/
void HLD_update(int u, int v, int d) {
    int tp1 = top[u], tp2 = top[v];
    while(tp1 != tp2) {
        if(dep[tp1] < dep[tp2]) {
            swap(u, v);
            swap(tp1, tp2);
        }
        update(id[tp1], id[u], d, 1, rear, 1);
        u = fa[tp1]; tp1 = top[u];
    }
    if(dep[u] > dep[v]) swap(u, v);
    update(id[u], id[v], d, 1, rear, 1);
}

int HLD_query(int u, int v) {
    int tp1 = top[u], tp2 = top[v], ret = 0;
    int lastla[2] = { -1, -1}, lastra[2] = { -1, -1}, la[2] = { -1, -1}, ra[2] = { -1, -1};
    while(tp1 != tp2) {
        if(dep[tp1] >= dep[tp2]) {
            ret += query(id[tp1], id[u], 1, rear, 1, la[0], ra[0]);
            if(ra[0] == lastla[0]) ret--;
            lastla[0] = la[0]; lastra[0] = ra[0];
            u = fa[tp1]; tp1 = top[u];
        } else {
            ret += query(id[tp2], id[v], 1, rear, 1, la[1], ra[1]);
            if(ra[1] == lastla[1]) ret--;
            lastla[1] = la[1]; lastra[1] = ra[1];
            v = fa[tp2]; tp2 = top[v];
        }
    }
    if(dep[u] <= dep[v]) {
        ret += query(id[u], id[v], 1, rear, 1, la[1], ra[1]);
        if(ra[1] == lastla[1]) ret--;
        if(la[1] == lastla[0]) ret--;
    } else {
        ret += query(id[v], id[u], 1, rear, 1, la[0], ra[0]);
        if(ra[0] == lastla[0]) ret--;
        if(la[0] == lastla[1]) ret--;
    }
    return ret;
}

int main() {
    int n, m; //FIN;
    while(~scanf("%d%d", &n, &m)) {
        edge_init();
        for(int i = 1; i <= n; i++) {
            scanf("%d", &B[i]);
        }
        for(int i = 1; i <= n - 1; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            edge_add(u, v);
            edge_add(v, u);
        }
        HLD_presolve();

        while(m--) {
            char op[10];
            int a, b, c;
            scanf("%s", op);
            if(op[0] == 'Q') {
                scanf("%d%d", &a, &b);
                printf("%d\n", HLD_query(a, b));
            } else {
                scanf("%d%d%d", &a, &b, &c);
                HLD_update(a, b, c);
            }
        }
    }
    return 0;
}