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  • hysbz 2243 染色(树链剖分)

    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,可是区间合并比較简单,节点仅仅要记录左右端点的颜色就可以。

    #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("
    ");
        */
    }
    
    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
    ", query(a, b));
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/gcczhongduan/p/5141992.html
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