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  • 2016湖南省赛 I Tree Intersection(线段树合并,树链剖分)

    2016湖南省赛 I Tree Intersection(线段树合并,树链剖分)

    传送门:https://ac.nowcoder.com/acm/contest/1112/I

    题意:

    给你一个n个结点的树,树上每个节点有自己的颜色

    问你删除第i条边后形成的两颗子树有多少个相同的颜色

    题解:

    树链剖分写法:

    对于每一种颜色来说,如果这个颜色是在单独的一颗子树中,那么就不会对其他的边产生贡献,所以我们单独对每一种颜色对边的贡献讨论,如果这个颜色只有一个,那么就不会产生贡献,否则,他就可以在两个相同颜色之间的路径上的边产生贡献,用树剖剖下来后区间加一加就可以了

    线段树合并的写法

    对于每一种颜色建权值线段树,线段树合并统计答案即可

    代码:

    树链剖分写法:

    /**
     *        ┏┓    ┏┓
     *        ┏┛┗━━━━━━━┛┗━━━┓
     *        ┃       ┃  
     *        ┃   ━    ┃
     *        ┃ >   < ┃
     *        ┃       ┃
     *        ┃... ⌒ ...  ┃
     *        ┃       ┃
     *        ┗━┓   ┏━┛
     *          ┃   ┃ Code is far away from bug with the animal protecting          
     *          ┃   ┃   神兽保佑,代码无bug
     *          ┃   ┃           
     *          ┃   ┃        
     *          ┃   ┃
     *          ┃   ┃           
     *          ┃   ┗━━━┓
     *          ┃       ┣┓
     *          ┃       ┏┛
     *          ┗┓┓┏━┳┓┏┛
     *           ┃┫┫ ┃┫┫
     *           ┗┻┛ ┗┻┛
     */
    // warm heart, wagging tail,and a smile just for you!
    //
    //                            _ooOoo_
    //                           o8888888o
    //                           88" . "88
    //                           (| -_- |)
    //                           O  =  /O
    //                        ____/`---'\____
    //                      .'  |     |//  `.
    //                     /  |||  :  |||//  
    //                    /  _||||| -:- |||||-  
    //                    |   |   -  /// |   |
    //                    | \_|  ''---/''  |   |
    //                      .-\__  `-`  ___/-. /
    //                  ___`. .'  /--.--  `. . __
    //               ."" '<  `.___\_<|>_/___.'  >'"".
    //              | | :  `- \`.;` _ /`;.`/ - ` : | |
    //                 `-.   \_ __ /__ _/   .-` /  /
    //         ======`-.____`-.___\_____/___.-`____.-'======
    //                            `=---='
    //        ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
    //                     佛祖保佑      永无BUG
    #include <set>
    #include <map>
    #include <stack>
    #include <cmath>
    #include <queue>
    #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("*********
    ")
    #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)<<"]
    "
    #define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]
    "
    #define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]
    "
    const int maxn = 3e5 + 5;
    const int INF = 0x3f3f3f3f;
    const int mod = 1e9 + 7;
    const double Pi = acos(-1);
    LL gcd(LL a, LL b) {
        return b ? gcd(b, a % b) : a;
    }
    LL lcm(LL a, LL b) {
        return a / gcd(a, b) * b;
    }
    double dpow(double a, LL b) {
        double ans = 1.0;
        while(b) {
            if(b % 2)ans = ans * a;
            a = a * a;
            b /= 2;
        } return ans;
    }
    LL quick_pow(LL x, LL y) {
        LL ans = 1;
        while(y) {
            if(y & 1) {
                ans = ans * x % mod;
            } x = x * x % mod;
            y >>= 1;
        } return ans;
    }
    int n;
    int a[maxn], sz[maxn], dep[maxn], fa[maxn], top[maxn], w[maxn], son[maxn], W[maxn];
    int sum[maxn << 2], lazy[maxn << 2], vis[maxn << 2];
     
    struct EDGE {
        int u, v, nt;
    } edge[maxn << 1];
    int head[maxn], summ, cnt;
    int u[maxn];
    int v[maxn];
    void add_edge(int u, int v) {
        edge[++summ].u = u;
        edge[summ].v = v;
        edge[summ].nt = head[u];
        head[u] = summ;
    }
     
    void dfs1(int u) {
        sz[u] = 1; son[u] = 0;
        for (int i = head[u]; ~i; i = edge[i].nt) {
            int v = edge[i].v;
            if (v != fa[u]) {
                fa[v] = u;
                dep[v] = dep[u] + 1;
                dfs1(v);
                sz[u] += sz[v];
                if (sz[v] > sz[son[u]]) son[u] = v;
            }
        }
    }
     
    void dfs2(int u, int tp, int x) {
        top[u] = tp; w[u] = ++cnt; W[cnt] = u;
        if (son[u]) dfs2(son[u], tp, 1);
        for (int i = head[u]; ~i; i = edge[i].nt) {
            int v = edge[i].v;
            if (v == son[u] || v == fa[u]) continue;
            dfs2(v, v, 2);
        }
    }
    vector<int> vec[maxn];
     
    void init() {
        memset(head, -1, sizeof(head));
        summ = 1; cnt = 0;
        for(int i = 1; i <= n; i++) {
            vec[i].clear();
        }
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            vec[a[i]].push_back(i);
        }
        for (int i = 1; i < n; i++) {
            // int u, v;
            scanf("%d %d", &u[i], &v[i]);
            add_edge(u[i], v[i]);
            add_edge(v[i], u[i]);
        }
        dep[1] = 1; fa[1] = 0;
        dfs1(1);
        dfs2(1, 1, 1);
    }
     
    void pushup(int rt) {
        if(vis[ls] == vis[rs]) vis[rt] = vis[ls];
        else vis[rt] = -1;
    }
     
    void pushdown(int rt, int mid) {
        if (lazy[rt]) {
            lazy[rt << 1] = lazy[rt];
            lazy[rt << 1 | 1] = lazy[rt];
            vis[rt << 1] = 0;
            vis[rt << 1 | 1] = 0;
            lazy[rt] = 0;
        }
    }
     
    void build(int l, int r, int rt) {
        lazy[rt] = 0;
        sum[rt] = 0;
        if (l == r) {
            vis[rt] = 0;
            return;
        }
        int mid = (l + r) >> 1;
        build(lson);
        build(rson);
    }
     
    void update(int L, int R, int val, int l, int r, int rt) {
        // debug2(L, R);
        if(L > R || r < L || R < l) return;
        if(vis[rt] == 1) return;
        if (L <= l && r <= R) {
            // sum[rt] += val * (r - l + 1);
            // lazy[rt] += val;
            if(vis[rt] == 0) {
                vis[rt] = 1;
                sum[rt]++;
                return;
            }
        }
        pushdown(rt, r - l + 1);
        int mid = (l + r) >> 1;
        if (L <= mid) update(L, R, val, lson);
        if (R > mid) update(L, R, val, rson);
        pushup(rt);
    }
     
    int query(int pos, int l, int r, int rt) {
        if (l == r) {
            return sum[rt];
        }
        pushdown(rt, r - l + 1);
        int mid = (l + r) >> 1;
        int ans = sum[rt];
        if (pos <= mid)  ans += query(pos, lson);
        else  ans += query(pos, rson);
        return ans;
    }
     
    int LCA(int x, int y) {
        // debug2(x, y);
        while (top[x] != top[y]) {
            // bug;
            if (dep[top[x]] < dep[top[y]]) swap(x, y);
            update(w[top[x]], w[x], 1, 1, n, 1);
            x = fa[top[x]];
        }
        if (dep[x] > dep[y])
            std::swap(x, y);
        update(w[x] + 1, w[y], 1, 1, n, 1);
        return x;
    }
     
    int main() {
    #ifndef ONLINE_JUDGE
        FIN
    #endif
        while(scanf("%d", &n) != EOF) {
            init();
            build(1, n, 1);
            // for(int i = 1; i <= n; i++) {
            //     // printf("%d
    ", w[]);
            //     // debug1(w[i]);
            // }
            for(int i = 1; i <= n; i++) {
                vis[1] = 0;
                lazy[1] = 1;
                int sz = vec[i].size();
                if(sz < 2) continue;
                int bb = LCA(vec[i][0], vec[i][1]);
                for(int j = 2; j < sz; j++) {
                    // debug2(vec[i][0], vec[i][1]);
                    bb = LCA(bb, vec[i][j]);
                }
            }
     
            for(int i = 1; i < n; i++) {
                int ans;
                // debug2(v[i], u[i]);
                if(fa[v[i]] == u[i]) {
     
                    ans = query(w[v[i]], 1, n, 1);
                } else {
     
                    ans = query(w[u[i]], 1, n, 1);
                }
                printf("%d
    ", ans);
            }
        }
     
        return 0;
    }
    

    线段树合并写法

    /**
     *        ┏┓    ┏┓
     *        ┏┛┗━━━━━━━┛┗━━━┓
     *        ┃       ┃  
     *        ┃   ━    ┃
     *        ┃ >   < ┃
     *        ┃       ┃
     *        ┃... ⌒ ...  ┃
     *        ┃       ┃
     *        ┗━┓   ┏━┛
     *          ┃   ┃ Code is far away from bug with the animal protecting          
     *          ┃   ┃   神兽保佑,代码无bug
     *          ┃   ┃           
     *          ┃   ┃        
     *          ┃   ┃
     *          ┃   ┃           
     *          ┃   ┗━━━┓
     *          ┃       ┣┓
     *          ┃       ┏┛
     *          ┗┓┓┏━┳┓┏┛
     *           ┃┫┫ ┃┫┫
     *           ┗┻┛ ┗┻┛
     */
    // warm heart, wagging tail,and a smile just for you!
    //
    //                            _ooOoo_
    //                           o8888888o
    //                           88" . "88
    //                           (| -_- |)
    //                           O  =  /O
    //                        ____/`---'\____
    //                      .'  |     |//  `.
    //                     /  |||  :  |||//  
    //                    /  _||||| -:- |||||-  
    //                    |   |   -  /// |   |
    //                    | \_|  ''---/''  |   |
    //                      .-\__  `-`  ___/-. /
    //                  ___`. .'  /--.--  `. . __
    //               ."" '<  `.___\_<|>_/___.'  >'"".
    //              | | :  `- \`.;` _ /`;.`/ - ` : | |
    //                 `-.   \_ __ /__ _/   .-` /  /
    //         ======`-.____`-.___\_____/___.-`____.-'======
    //                            `=---='
    //        ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
    //                     佛祖保佑      永无BUG
    #include <set>
    #include <map>
    #include <stack>
    #include <cmath>
    #include <queue>
    #include <cstdio>
    #include <string>
    #include <bitset>
    #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("*********
    ")
    #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)<<"]
    "
    #define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]
    "
    #define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]
    "
    const int maxn = 1e5 + 5;
    const int INF = 0x3f3f3f3f;
    const int mod = 1e9 + 7;
    const double Pi = acos(-1);
    LL gcd(LL a, LL b) {
        return b ? gcd(b, a % b) : a;
    }
    LL lcm(LL a, LL b) {
        return a / gcd(a, b) * b;
    }
    double dpow(double a, LL b) {
        double ans = 1.0;
        while(b) {
            if(b % 2)ans = ans * a;
            a = a * a;
            b /= 2;
        } return ans;
    }
    LL quick_pow(LL x, LL y) {
        LL ans = 1;
        while(y) {
            if(y & 1) {
                ans = ans * x % mod;
            } x = x * x % mod;
            y >>= 1;
        } return ans;
    }
    int col[maxn];
    int cnt[maxn];
    struct EDGE {
        int v, nxt;
    } edge[maxn << 1];
    int head[maxn], tot;
    void init() {
        memset(head, -1, sizeof(head));
        tot = 0;
    }
    void add_edge(int u, int v) {
        edge[tot].v = v;
        edge[tot].nxt = head[u];
        head[u] = tot++;
    }
     
    struct node {
        int l, r, sum, val;
    } tree[maxn * 20];
    int root[maxn];
    int rt_cnt = 0;
    int n;
    int ans[maxn];
    void push_up(int rt) {
        tree[rt].sum = tree[tree[rt].l].sum + tree[tree[rt].r].sum;
    }
    int build(int l, int r, int pos) {
        int rt = ++rt_cnt;
        tree[rt].l = 0;
        tree[rt].r = 0;
        tree[rt].sum = 0;
        if(l == r) {
            tree[rt].val = 1;
            if(tree[rt].val != cnt[l]) {
                tree[rt].sum = 1;
            } else {
                tree[rt].sum = 0;
            }
            return rt;
        }
        int mid = (l + r) >> 1;
        if(pos <= mid) tree[rt].l = build(l, mid, pos);
        else tree[rt].r = build(mid + 1, r, pos);
        push_up(rt);
        return rt;
    }
    void merge(int &x, int y, int l, int r) {
        if(!x || !y) {
            if(!x) x = y;
            return;
        }
     
        if(l == r) {
            tree[x].val += tree[y].val;
            tree[x].sum = (tree[x].val != cnt[l]);
            return;
        }
        int mid = (l + r) >> 1;
        merge(tree[x].l, tree[y].l, l, mid);
        merge(tree[x].r, tree[y].r, mid + 1, r);
        push_up(x);
    }
     
    void dfs(int u, int fa, int id) {
        root[u] = build(1, n, col[u]);
        for(int i = head[u]; i != -1; i = edge[i].nxt) {
            int v = edge[i].v;
            if(v == fa) continue;
            dfs(v, u, i);
            merge(root[u], root[v], 1, n);
        }
        if(u != 1) {
            int tid = id / 2 + 1;
            ans[tid] = tree[root[u]].sum;
        }
    }
    int main() {
    #ifndef ONLINE_JUDGE
        FIN
    #endif
     
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &col[i]);
            // vec[col[i]].push_back(i);
            cnt[col[i]]++;
        }
        init();
        for(int i = 1, u, v; i < n; i++) {
            scanf("%d%d", &u, &v);
            add_edge(u, v);
            add_edge(v, u);
        }
        // debug1(n);
        dfs(1, -1, -1);
     
        for(int i = 1; i <= n - 1; i++) {
            printf("%d
    ", ans[i]);
        }
     
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/buerdepepeqi/p/11650725.html
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