hdu 3966 Aragorn's Story
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树链剖分的入门题。
题目大意:n个结点的树,每个结点有一个值。两种操作:1、将结点a-b路径上的结点值都加上或减去一个值。 2、询问结点a的值。
建树之后,将树状数组初始化为0,假设每个结点的值也都为0。然后进行更新。
设结点i建树之后的序号为id[i],初始值为A[i],则在id[i]位置增加A[i],在id[i]+1的位置减少A[i],这样在询问结点的值时,只需用树状数组的求一下前id[i]项和即为序号为id[i]的结点的值。
对于更新操作,对于两个结点u、v。找到其所在链的顶端结点top[u],top[v],比较其深度,若dep[top[u]]>dep[top[v],则将结点序号为id[top[u]]到id[u]的值更新,再令u=fa[top[u]。直到top[u]=top[v],此时若u!=v,则再更新u-v路径上结点的值。
注意:为防止爆栈,得加上这句#pragma comment(linker,"/STACK:102400000,102400000")
#include <cstdio>#include <cstring>#include <vector>#include <algorithm>#pragma comment(linker,"/STACK:102400000,102400000")using namespace std;#define maxn 50005int A[maxn],dep[maxn], fa[maxn], son[maxn], num[maxn], top[maxn], id[maxn], C[maxn],tot;vector<int> g[maxn];int lowbit(int x){return x&(-x);}void dfs1(int u){ num[u] = 1,son[u] = 0; for (int i = 0; i < g[u].size(); i++) if (g[u][i] != fa[u]) { fa[g[u][i]]=u; dep[g[u][i]]=dep[u]+1; dfs1(g[u][i]); num[u] += num[g[u][i]]; if (num[son[u]] < num[g[u][i]]) son[u] = g[u][i]; }}void dfs2(int u, int tp){ id[u] = ++tot; top[u] = tp; if (son[u]) dfs2(son[u], tp); for (int i = 0; i < g[u].size(); i++) { int v = g[u][i]; if (v == fa[u] || v == son[u]) continue; dfs2(v, v); }}void add(int x, int v){ while (x < maxn) { C[x] += v; x += lowbit(x); }}void add(int l, int r, int v){ add(l, v); add(r + 1, -v);}void update(int u, int v, int w){ int f1 = top[u], f2 = top[v]; while (f1 != f2) { if (dep[f1] < dep[f2]) { swap(f1, f2); swap(u, v); } add(id[f1], id[u], w); u = fa[f1]; f1 = top[u]; } if (dep[u] > dep[v]) swap(u, v); add(id[u], id[v], w);}int query(int x){ int ans = 0; while (x) { ans += C[x]; x -= lowbit(x); } return ans;}int main(){ int n,m,p; while (~scanf("%d%d%d", &n, &m, &p)) { memset(C, 0, sizeof(C)); memset(num,0,sizeof(num)); dep[1]=fa[1]=tot=0; for (int i = 1; i <= n; i++) { scanf("%d", &A[i]); g[i].clear(); } int u, v; while (m--) { scanf("%d%d", &u, &v); g[u].push_back(v); g[v].push_back(u); } dfs1(1); dfs2(1, 1); for (int i = 1; i <= n; i++) add(id[i], id[i], A[i]); char q[2]; int a, b, c; while (p--) { scanf("%s", q); if (q[0] == 'I' || q[0] == 'D') { scanf("%d%d%d", &a, &b, &c); if (q[0] == 'D') c = -c; update(a, b, c); } else { scanf("%d", &a); printf("%d\n", query(id[a])); } } } return 0;}
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