• HYSBZ 2243-染色 (树链剖分)


    1A!!! 哈哈哈哈哈没看题解 没套模板哈哈哈哈 太感动了!!

    如果只是线段树的话这道题倒是不难,只要记录左右边界就好了,类似很久以前做的hotel的题

    但是树上相邻的段会有连续的

    树上top[x]和fa[top[x]]是连续的,但是线段树上是算不到的,所以要判断下

    线段树记录的是区间的数量,但是求单点的时候求得是颜色,需要注意下

    #include <iostream>
    #include <cstring>
    #include <cstdio>
    
    using namespace std;
    const int N = 100005;
    int a[N];
    
    struct Edge {
        int to, next;
    } edge[N*2];
    int head[N], cntE;
    
    void addedge(int u, int v) {
        edge[cntE].to = v; edge[cntE].next = head[u]; head[u] = cntE++;
        edge[cntE].to = u; edge[cntE].next = head[v]; head[v] = cntE++;
    }
    
    int dep[N], fa[N], sz[N], son[N];
    void dfs1(int u, int pre, int d) {
        dep[u] = d;
        sz[u] = 1;
        fa[u] = pre;
        for (int i = head[u]; ~i; i = edge[i].next) {
            int v = edge[i].to;
            if (v != pre) {
                dfs1(v, u, d+1);
                sz[u] += sz[v];
                if (son[u] == -1 || sz[v] > sz[son[u]]) son[u] = v;
            }
        }
    }
    
    int top[N], dfn[N], rk[N], idx;
    void dfs2(int u, int tp) {
        top[u] = tp;
        dfn[u] = ++idx;
        rk[idx] = u;
        if (son[u] == -1) return ;
        dfs2(son[u], tp);
        for (int i = head[u]; ~i; i = edge[i].next) {
            int v = edge[i].to;
            if (v != fa[u] && v != son[u]) {
                dfs2(v, v);
            }
        }
    }
    
    int tr[N<<2], ltr[N<<2], rtr[N<<2]; // tr[i] means the number of color segment
    int fg[N<<2];
    #define lson o<<1
    #define rson o<<1|1
    
    void pushup(int o) {
        ltr[o] = ltr[lson];
        rtr[o] = rtr[rson];
        tr[o] = tr[lson] + tr[rson];
        if (rtr[lson] == ltr[rson]) tr[o]--;
    }
    
    void pushdown(int o) {
        if (fg[o]) {
            tr[lson] = tr[rson] = 1;
            fg[lson] = fg[rson] = fg[o];
            ltr[lson] = ltr[rson] = ltr[o];
            rtr[lson] = rtr[rson] = rtr[o];
            fg[o] = 0;
        }
    }
    
    void build(int o, int l, int r) {
        fg[o] = 0;
        if (l == r) {
            tr[o] = 1;
            ltr[o] = rtr[o] = a[rk[l]];
            return;
        }
        int mid = (l+r) >> 1;
        build(lson, l, mid);
        build(rson, mid+1, r);
        pushup(o);
    }
    
    void change(int o, int l, int r, int L, int R, int v) {
        if (l >= L && r <= R) {
            fg[o] = 1;
            tr[o] = 1;
            ltr[o] = rtr[o] = v;
            return ;
        }
        pushdown(o);
        int mid = (l+r) >> 1;
        if (mid >= L) change(lson, l, mid, L, R, v);
        if (mid < R) change(rson, mid+1, r, L, R, v);
        pushup(o);
    }
    
    void CHANGE(int x, int y, int n, int c) {
        while (top[x] != top[y]) {
            if (dep[top[x]] < dep[top[y]]) swap(x, y);
            change(1, 1, n, dfn[top[x]], dfn[x], c);
            x = fa[top[x]];
        }
        if (dep[x] > dep[y]) swap(x, y);
        change(1, 1, n, dfn[x], dfn[y], c);
    }
    
    int query(int o, int l, int r, int L, int R) {
        if (l >= L && r <= R) return tr[o];
        pushdown(o);
        int mid = (l+r) >> 1;
        if (mid < L) {
            return query(rson, mid+1, r, L, R);
        } else if (mid >= R) {
            return query(lson, l, mid, L, R);
        } else {
            int ans = query(lson, l, mid, L, R);
            ans += query(rson, mid+1, r, L, R);
            if (ltr[rson] == rtr[lson]) ans--;
            return ans;
        }
    }
    
    int qq(int o, int l, int r, int p) {
        if (l == r) return ltr[o];
        pushdown(o);
        int mid = (l+r) >> 1;
        if (mid >= p) return qq(lson, l, mid, p);
        return qq(rson, mid+1, r, p);
    }
    
    int QUERY(int x, int y, int n) {
        int ans = 0;
        while (top[x] != top[y]) {
            if (dep[top[x]] < dep[top[y]]) swap(x, y);
            ans += query(1, 1, n, dfn[top[x]], dfn[x]);
            if (qq(1, 1, n, dfn[top[x]]) == qq(1, 1, n, dfn[fa[top[x]]])) --ans;
            x = fa[top[x]];
        }
        if (dep[x] > dep[y]) swap(x, y);
        ans += query(1, 1, n, dfn[x], dfn[y]);
        return ans;
    }
    
    void init() {
        memset(head, -1, sizeof head);
        memset(son, -1, sizeof son);
        idx = cntE = 0;
    }
    
    int main()
    {
        //freopen("in.txt", "r", stdin);
        int n, m;
        while (~scanf("%d%d", &n, &m)) {
            init();
            for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
            int u, v, c;
            for (int i = 1; i < n; ++i) {
                scanf("%d%d", &u, &v);
                addedge(u, v);
            }
            dfs1(1, 0, 0); dfs2(1, 1); build(1, 1, n);
    
            char op[10];
            while (m--) {
                scanf("%s", op);
                if (*op == 'Q') {
                    scanf("%d%d", &u, &v);
                    printf("%d
    ", QUERY(u, v, n));
                } else {
                    scanf("%d%d%d", &u, &v, &c);
                    CHANGE(u, v, n, c);
                }
            }
        }
        return 0;
    }
  • 相关阅读:
    一文带你认知定时消息发布RocketMQ
    Python图像处理丨基于KMeans聚类的图像区域分割
    openGauss内核分析:SQL by pass & 经典执行器
    手把手教你君正X2000开发板的OpenHarmony环境搭建
    中秋节,华为云AI送上超级大月亮制作教程,体验赢开发者键鼠套装
    强扩展、强一致、高可用…GaussDB成为游戏行业的心头爱
    测试工程师面试指南(二) | 大咖公开课
    该如何测客户端专项测试?
    测试冲榜获奖名单公布,来领 Kindle、小米蓝牙耳机、测试好书!
    【测开基础之计算机网络】二: 物理层_网络_TesterAllen的博客CSDN博客
  • 原文地址:https://www.cnblogs.com/wenruo/p/5927543.html
Copyright © 2020-2023  润新知