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  • FZU 2082 过路费(树链剖分)

    FZU 2082 过路费

    题目链接

    树链抛分改动边的模板题

    代码:

    #include <cstdio>
    #include <cstring>
    #include <vector>
    using namespace std;
    
    typedef long long ll;
    const int N = 50005;
    
    int dep[N], id[N], sz[N], top[N], son[N], fa[N], idx;
    int n, m;
    ll bit[N];
    
    struct Edge {
    	int u, v;
    	ll val;
    	void read() {
    		scanf("%d%d%lld", &u, &v, &val);
    	}
    } e[N];
    
    vector<int> g[N];
    
    inline int lowbit(int x) {
    	return x&(-x);
    }
    
    void dfs1(int u, int f, int d) {
    	dep[u] = d;
    	sz[u] = 1;
    	fa[u] = f;
    	son[u] = 0;
    	for (int i = 0; i < g[u].size(); i++) {
    		int v = g[u][i];
    		if (v == f) continue;
    		dfs1(v, u, d + 1);
    		sz[u] += sz[v];
    		if (sz[son[u]] < sz[v])
    			son[u] = v;
    	}
    }
    
    void dfs2(int u, int tp) {
    	top[u] = tp;
    	id[u] = idx++;
    	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, ll v) {
    	while (x < N) {
    		bit[x] += v;
    		x += lowbit(x);
    	}
    }
    
    ll query(int x) {
    	ll ans = 0;
    	while (x) {
    		ans += bit[x];
    		x -= lowbit(x);
    	}
    	return ans;
    }
    
    ll query(int l, int r) {
    	return query(r) - query(l - 1);
    }
    
    ll gao(int u, int v) {
    	int tp1 = top[u], tp2 = top[v];
    	ll ans = 0;
    	while (tp1 != tp2) {
    		if (dep[tp1] < dep[tp2]) {
    			swap(tp1, tp2);
    			swap(u, v);
    		}
    		ans += query(id[tp1], id[u]);
    		u = fa[tp1];
    		tp1 = top[u];
    	}
    	if (u == v) return ans;
    	if (dep[u] > dep[v]) swap(u, v);
    	ans += query(id[son[u]], id[v]);
    	return ans;
    }
    
    int main() {
    	while (~scanf("%d%d", &n, &m)) {
    		idx = 0;
    		memset(bit, 0, sizeof(bit));
    		for (int i = 1; i <= n; i++) g[i].clear();
    		for (int i = 1; i < n; i++) {
    			e[i].read();
    			g[e[i].u].push_back(e[i].v);
    			g[e[i].v].push_back(e[i].u);
    		}
    		dfs1(1, 0, 1);
    		dfs2(1, 1);
    		for (int i = 1; i < n; i++) {
    			if (dep[e[i].u] < dep[e[i].v]) swap(e[i].u, e[i].v);
    			add(id[e[i].u], e[i].val);
    		}
    		int ty, a, b;
    		while (m--) {
    			scanf("%d%d%d", &ty, &a, &b);
    			if (ty == 0) {
    				ll tmp = query(id[e[a].u], id[e[a].u]);
    				add(id[e[a].u], (ll)b - tmp);
    			} else printf("%lld
    ", gao(a, b));
    		}
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/mengfanrong/p/4090120.html
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