[CF1076E]Vasya and a Tree

题目大意：给定一棵以$1$为根的树，$m$次操作，第$i$次为对以$v_i$为根的深度小于等于$d_i$的子树的所有节点权值加$x_i$。最后输出每个节点的值

题解：可以把操作离线，每次开始遍历到一个节点，把以它为根的操作加上，结束时把这个点的操作删去。

因为是$dfs$，所以一个点对同一深度的贡献是一定的，可以用树状数组区间加减，求前缀和。但是因为是$dfs$，完全可以把前缀和传入，就不需要树状数组了。

卡点：无

C++ Code：

#include <cstdio>
#include <vector>
#include <cctype>
namespace __IO {namespace R {int x, ch;inline int read() {ch = getchar();while (isspace(ch)) ch = getchar();for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);return x;}}
}
using __IO::R::read;#define maxn 300010int head[maxn], cnt;
struct Edge {int to, nxt;
} e[maxn << 1];
inline void add(int a, int b) {e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
}int n, m;
struct Modify {int d, x;inline Modify() {}inline Modify(int __d, int __x) :d(__d), x(__x){}
};
std::vector<Modify> S[maxn];long long pre[maxn], ans[maxn];
void dfs(int u, int fa = 0, int dep = 0, long long sum = 0) {for (std::vector<Modify>::iterator it = S[u].begin(); it != S[u].end(); it++) {pre[dep] += it -> x;if (dep + it -> d + 1 <= n) pre[dep + it -> d + 1] -= it -> x;}sum += pre[dep];ans[u] = sum;for (int i = head[u]; i; i = e[i].nxt) {int v = e[i].to;if (v != fa) dfs(v, u, dep + 1, sum);}for (std::vector<Modify>::iterator it = S[u].begin(); it != S[u].end(); it++) {pre[dep] -= it -> x;if (dep + it -> d + 1 <= n) pre[dep + it -> d + 1] += it -> x;}
}
int main() {n = read();for (int i = 1, a, b; i < n; i++) {a = read(), b = read();add(a, b);add(b, a);}m = read();for (int i = 1, v, d, x; i <= m; i++) {v = read(), d = read(), x = read();S[v].push_back(Modify(d, x));}dfs(1);for (int i = 1; i <= n; i++) {printf("%lld", ans[i]);putchar(i == n ? '\n' : ' ');}return 0;
}

转载于:https://www.cnblogs.com/Memory-of-winter/p/9995403.html

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