hysbz 2243 染色(树链剖分)

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题目链接：hysbz 2243 染色

题目大意：略。

解题思路：树链剖分+线段树的区间合并，但是区间合并比较简单，节点只要记录左右端点的颜色即可。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 1e5 + 5;

int N, M, ne, val[maxn], first[maxn], jump[maxn * 2];
int id, far[maxn], son[maxn], dep[maxn], top[maxn], cnt[maxn], idx[maxn];
struct Edge {
    int u, v;
    void set (int u, int v) {
        this->u = u;
        this->v = v;
    }
}ed[maxn * 2];

inline void add_Edge(int a, int b) {
    ed[ne].set(a, b);
    jump[ne] = first[a];
    first[a] = ne++;
}

void dfs_fir(int u, int pre, int d) {
    far[u] = pre;
    dep[u] = d;
    son[u] = 0;
    cnt[u] = 1;

    for (int i = first[u]; i + 1; i = jump[i]) {
        int k = ed[i].v;
        if (k == pre)
            continue;
        dfs_fir(k, u, d + 1);
        cnt[u] += cnt[k];
        if (cnt[son[u]] < cnt[k])
            son[u] = k;
    }
}

void dfs_sec(int u, int rot) {
    top[u] = rot;
    idx[u] = id++;

    if (son[u])
        dfs_sec(son[u], rot);

    for (int i = first[u]; i + 1; i = jump[i]) {
        int k = ed[i].v;
        if (k == far[u] || k == son[u])
            continue;
        dfs_sec(k, k);
    }
}

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], v[maxn << 2];
struct Seg {
    int l, r, s;
    Seg (int s = 0, int l = 0, int r = 0) {
        this->l = l;
        this->r = r;
        this->s = s;
    }
    void maintain(int d) {
        s = 1;
        l = r = d;
    }
}nd[maxn << 2];

inline Seg merge(const Seg& L, const Seg& R) {
    if (L.s == 0) return R;
    if (R.s == 0) return L;
    return Seg(L.s + R.s + (L.r == R.l ? -1 : 0), L.l, R.r);
}

inline void pushdown(int u) {
    if (v[u] != -1) {
        v[lson(u)] = v[rson(u)] = v[u];
        nd[lson(u)].maintain(v[u]);
        nd[rson(u)].maintain(v[u]);
        v[u] = -1;
    }
}

inline void pushup(int u) {
    nd[u] = merge(nd[lson(u)], nd[rson(u)]);
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    v[u] = -1;

    if (l == r) {
        nd[u].maintain(-1);
        return;
    }

    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    pushup(u);
}

void modify (int u, int l, int r, int w) {
    if (l <= lc[u] && rc[u] <= r) {
        v[u] = w;
        nd[u].maintain(w);
        return;
    }

    pushdown(u);
    int mid = (lc[u] + rc[u]) / 2;
    if (l <= mid)
        modify(lson(u), l, r, w);
    if (r > mid)
        modify(rson(u), l, r, w);
    pushup(u);
}

Seg query(int u, int l, int r) {
    if (l <= lc[u] && rc[u] <= r)
        return nd[u];

    pushdown(u);
    Seg ret;
    int mid = (lc[u] + rc[u]) / 2;
    if (l <= mid)
        ret = merge(ret, query(lson(u), l, r));
    if (r > mid)
        ret = merge(ret, query(rson(u), l, r));
    pushup(u);
    return ret;
}

void init () {
    ne = 0;
    id = 1;
    memset(first, -1, sizeof(first));

    for (int i = 1; i <= N; i++)
        scanf("%d", &val[i]);

    int a, b;
    for (int i = 1; i < N; i++) {
        scanf("%d%d", &a, &b);
        add_Edge(a, b);
        add_Edge(b, a);
    }
    dfs_fir(1, 0, 0);
    dfs_sec(1, 1);

    build(1, 1, N);
    for (int i = 1; i <= N; i++)
        modify(1, idx[i], idx[i], val[i]);

    /*
    for (int i = 1; i <= N; i++) {
        Seg ret = query(1, idx[i], idx[i]);
        printf("%d ", ret.r);
    }
    printf("\n");
    */
}

void modify(int a, int b, int w) {
    int p = top[a], q = top[b];
    while (p != q) {
        if (dep[p] < dep[q]) {
            swap(p, q);
            swap(a, b);
        }
        modify(1, idx[p], idx[a], w);
        a = far[p];
        p = top[a];
    }
    if (dep[a] > dep[b])
        swap(a, b);
    modify(1, idx[a], idx[b], w);
}

int query(int a, int b) {
    int p = top[a], q = top[b];
    Seg one, two;
    while (p != q) {
        if (dep[p] < dep[q]) {
            swap(p, q);
            swap(a, b);
            swap(one, two);
        }
        one = merge(query(1, idx[p], idx[a]), one);
        a = far[p];
        p = top[a];
    }

    if (dep[a] > dep[b]) {
        swap(a, b);
        swap(one, two);
    }

    two = merge(query(1, idx[a], idx[b]), two);
    int ret = one.s + two.s;
    if (one.l == two.l) ret--;
    return ret;
}

int main () {

    while (scanf("%d%d", &N, &M) == 2) {
        init();

        int a, b, w;
        char op[5];
        while (M--) {
            scanf("%s%d%d", op, &a, &b);
            if (op[0] == ‘C‘) {
                scanf("%d", &w);
                modify(a, b, w);
            } else
                printf("%d\n", query(a, b));
        }
    }
    return 0;
}

hysbz 2243 染色(树链剖分)

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/40181073