fzu 2277 Change [第八届福建省大学生程序设计竞赛 Problem F] [线段树]

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题意: 给定一棵根为1, n个结点的树. 有q个操作,有两种不同的操作

(1) 1 v k x : a[v] += x, a[v '] += x – k(v '为v的儿子), a[v ' '] += x – 2 * k(v ' '是v '的儿子) ... ;

(2) 2 v : 输出a[v] % (1e9 + 7);

 

分析: 先dfs遍历, 得到这棵树的dfs序列, 并且记录每个结点的子孙(包括自己)的起始和结束位置, 以及每个结点的深度. 则对于1操作: v的每个子孙结点的变化情况为a[i] += x – (deep[i] – deep[v])* k; 其中x + k * deep[v]对所有结点贡献相同,而deep[i] * k对每个结点的贡献度取决于结点i的深度. 所以可以用线段树, 对于每次操作一,所有v的子孙结点都加上x + k * deep[v], 然后对于deep[i] * k每次记录k, 到最后更新的时候再乘以结点的深度即可.

 

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>typedef long long ll;const int maxn = 3 * 1e5 + 10;const ll mod = 1e9 + 7;using namespace std;vector<int> G[maxn];int node[maxn], id1[maxn], id2[maxn], dep[maxn];ll xx[maxn << 2], kk[maxn << 2];int k, n;int Scan(){    int res = 0, ch, flag = 0;    if((ch = getchar()) == '-')        flag = 1;    else if(ch >= '0' && ch <= '9')        res = ch - '0';    while((ch = getchar()) >= '0' && ch <= '9' )        res = res * 10 + ch - '0';    return flag ? -res : res;}ll Scan_l(){    ll res = 0;    int ch, flag = 0;    if((ch = getchar()) == '-')        flag = 1;    else if(ch >= '0' && ch <= '9')        res = ch - '0';    while((ch = getchar()) >= '0' && ch <= '9' )        res = res * 10 + ch - '0';    return flag ? -res : res;}void dfs(int u, int fa, int d) {    node[k] = u;    id1[u] = id2[u] = k;    dep[k++] = d;    for(int i = 0; i < G[u].size(); i++) {        ll v = G[u][i];        if(v == fa) continue;        dfs(v, u, d + 1);    }    id2[u] = k - 1;}void pushdown(int rt) {    xx[rt << 1] += xx[rt];    xx[rt << 1] %= mod;    xx[rt << 1 | 1] += xx[rt];    xx[rt << 1 | 1] %= mod;    kk[rt << 1] += kk[rt];    kk[rt << 1] %= mod;    kk[rt << 1 | 1] += kk[rt];    kk[rt << 1 | 1] %= mod;    xx[rt] = 0; kk[rt] = 0;}void update(int L, int R, int l, int r, int rt, ll c, ll d) {    if(L <= l && r <= R) {        xx[rt] = ((xx[rt] + c) % mod + mod) % mod;        kk[rt] = ((kk[rt] + d) % mod + mod) % mod;        return ;    }    pushdown(rt);    int m = (l + r) / 2;    if(L <= m) update(L, R, l, m, rt << 1, c, d);    if(m < R) update(L, R, m + 1, r, rt << 1 | 1, c, d);}ll query(int L, int R, int l, int r, int rt) {    if(L <= l && r <= R) {        return xx[rt] + dep[L] * kk[rt];    }    pushdown(rt);    int m = (l + r) / 2;    if(R <= m) return query(L, R, l, m, rt << 1);    else return query(L, R, m + 1, r, rt << 1 | 1);}int main() {    int t, g, q;    t = Scan();    while(t--) {        k = 1;        memset(xx, 0, sizeof xx);        memset(kk, 0, sizeof kk);        for(int i = 0; i < maxn; i++) G[i].clear();        n = Scan();        for(int i = 2; i <= n; i++) {            g = Scan();            G[g].push_back(i);        }        dfs(1, 0, 1);        q = Scan();        while(q--) {            int op, root;            ll x, y;            op = Scan();            if(op == 1) {                root = Scan();                x = Scan_l();                y = Scan_l();                update(id1[root], id2[root], 1, n, 1, x + dep[id1[root]] * y, -y);            }            if(op == 2) {                root = Scan();                printf("%I64d\n", (query(id1[root], id1[root], 1, n, 1) % mod + mod) % mod);            }        }    }    return 0;}