• Luogu P3178 树上操作(树链剖分+线段树)


    题意

    见原题

    题解

    重链剖分模板题

    #include <cstdio>
    #include <algorithm>
    using std::swap;
    typedef long long ll;
    
    const int N = 1e5 + 10;
    int n, m, c[N], opt, x, y;
    int dep[N], siz[N], fa[N], son[N];
    int top[N], dfn[N], w[N], time;
    int cnt, from[N], to[N << 1], nxt[N << 1];
    ll val[N << 2], add[N << 2];
    inline void addEdge(int u, int v){
    	to[++cnt] = v, nxt[cnt] = from[u], from[u] = cnt;
    }
    
    void dfs1(int u) {
    	dep[u] = dep[fa[u]] + 1, siz[u] = 1;
    	for (int i = from[u]; i; i = nxt[i]) {
    		int v = to[i]; if(v == fa[u]) continue;
    		fa[v] = u, dfs1(v), siz[u] += siz[v];
    		if(siz[v] > siz[son[u]]) son[u] = v;
    	}
    }
    void dfs2(int u, int t) {
    	top[u] = t, dfn[u] = ++time, w[time] = c[u];
    	if(!son[u]) return ; dfs2(son[u], t);
    	for(int i = from[u]; i; i = nxt[i]) {
    		int v = to[i];
    		if(v != fa[u] && v != son[u])
    			dfs2(v, v);
    	}
    }
    
    inline void pushup(int o, int lc, int rc) {
    	val[o] = val[lc] + val[rc];
    }
    inline void pushdown(int o, int lc, int rc, int len) {
    	if(add[o]) {
    		add[lc] += add[o], add[rc] += add[o];
    		val[lc] += add[o] * (len - (len >> 1));
    		val[rc] += add[o] * (len >> 1);
    		add[o] = 0;
    	}
    }
    void build(int o = 1, int l = 1, int r = n) {
    	if(l == r) { val[o] = w[l]; return ; }
    	int mid = (l + r) >> 1, lc = o << 1, rc = lc | 1;
    	build(lc, l, mid), build(rc, mid + 1, r), pushup(o, lc, rc);
    }
    void upt(int ul, int ur, ll k, int o = 1, int l = 1, int r = n) {
    	if (l >= ul && r <= ur) {
    		add[o] += k, val[o] += k * (r - l + 1);
    		return ;
    	}
    	int mid = (l + r) >> 1, lc = o << 1, rc = lc | 1;
    	pushdown(o, lc, rc, r - l + 1);
    	if(ul <= mid) upt(ul, ur, k, lc, l, mid);
    	if(ur > mid) upt(ul, ur, k, rc, mid + 1, r);
    	pushup(o, lc, rc);
    }
    ll que(int ql, int qr, int o = 1, int l = 1, int r = n) {
    	if (l >= ql && r <= qr) return val[o];
    	int mid = (l + r) >> 1, lc = o << 1, rc = lc | 1; ll ret = 0;
    	pushdown(o, lc, rc, r - l + 1);
    	if(ql <= mid) ret = que(ql, qr, lc, l, mid);
    	if(qr > mid) ret += que(ql, qr, rc, mid + 1, r);
    	return ret;
    }
    
    ll sum(int x) {
    	int fx = top[x]; ll ret = 0;
    	while (fx != 1) ret += que(dfn[fx], dfn[x]), x = fa[fx], fx = top[x];
    	return ret + que(1, dfn[x]);
    }
    
    int main () {
    	scanf("%d%d", &n, &m);
    	for (int i = 1; i <= n; ++i) scanf("%d", c + i);
    	for (int i = 1, u, v; i < n; ++i) {
    		scanf("%d%d", &u, &v);
    		addEdge(u, v), addEdge(v, u);
    	}
    	dfs1(1), dfs2(1, 1), build();
    	while(m--) {
    		scanf("%d%d", &opt, &x);
    		if (opt == 3) printf("%lld
    ", sum(x));
    		else {
    			scanf("%d", &y);
    			if (opt == 1) upt(dfn[x], dfn[x], y);
    			else upt(dfn[x], dfn[x] + siz[x] - 1, y);
    		}
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/water-mi/p/9826131.html
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