POJ 2763 Housewife Wind(树链剖分+线段树)

解题思路

本题只有两种操作:求树上两点间路径的权值和、修改边权。

于是使用树链剖分就很简单了。

参考代码

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#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1

const int MAXN = 100001;

struct edge_orig {
	int u, v;
	int w;
} edge[MAXN];

struct edge {
	int to;
	int next;
} e[MAXN<<1];

int n, q, s, a, b, op;
int fa[MAXN], dep[MAXN], sz[MAXN], son[MAXN];
int idx, id[MAXN], top[MAXN], val[MAXN];
int head[MAXN], cnt;
int SUM[MAXN<<2];

void Add(int u, int v) {
	e[cnt].to = v;
	e[cnt].next = head[u];
	head[u] = cnt++;
}

void DFS1(int u, int pre, int d) {
	dep[u] = d;
	sz[u] = 1;
	son[u] = 0;
	fa[u] = pre;
	for(int i=head[u]; ~i; i=e[i].next) {
		int v = e[i].to;
		if(v != pre) {
			DFS1(v, u, d+1);
			sz[u] += sz[v];
			if(sz[v] > sz[son[u]])
				son[u] = v;
		}
	}
}

void DFS2(int u, int t) {
	top[u] = t;
	id[u] = ++idx;
	if(son[u]) DFS2(son[u], t);
	for(int i=head[u]; ~i; i=e[i].next) {
		int v = e[i].to;
		if(v!=fa[u] && v!=son[u])
			DFS2(v, v);
	}
}

void PushUp(int rt) {
	SUM[rt] = SUM[rt<<1] + SUM[rt<<1|1];
}

void Build(int l, int r, int rt) {
	if(l == r) {
		SUM[rt] = val[l];
		return;
	}
	int m = (l + r) >> 1;
	Build(lson);
	Build(rson);
	PushUp(rt);
}

void Update(int p, int v, int l, int r, int rt) {
	if(l == r) {
		SUM[rt] = v;
		return;
	}
	int m = (l + r) >> 1;
	if(p <= m) Update(p, v,lson);
	else Update(p, v, rson);
	PushUp(rt);
}

int Query(int L, int R, int l, int r, int rt) {
	if(L<=l && r<=R) return SUM[rt];
	int m = (l + r) >> 1;
	int ret = 0;
	if(L <= m) ret += Query(L, R, lson);
	if(R > m) ret += Query(L, R, rson);

	return ret;
}

int GetAns(int u, int v) {
	int ans = 0;
	while(top[u] != top[v]) {
		if(dep[top[u]] < dep[top[v]])
			swap(u, v);
		ans += Query(id[top[u]], id[u], 1, n, 1);
		u = fa[top[u]];
	}
	if(u == v) return ans;
	if(dep[u] > dep[v]) swap(u, v);
	ans += Query(id[son[u]], id[v], 1, n, 1);

	return ans;
}

int main(int argc, char const *argv[]) {
	memset(head, -1, sizeof head);
	scanf("%d %d %d", &n, &q, &s);
	for(int i=1; i<=n-1; ++i) {
		scanf("%d %d %d", &edge[i].u, &edge[i].v, &edge[i].w);
		Add(edge[i].u, edge[i].v);
		Add(edge[i].v, edge[i].u);
	}
	idx = 0;
	DFS1(1, -1, 1);
	DFS2(1, 1);
	for(int i=1; i<=n-1; ++i) {
		if(dep[edge[i].u] < dep[edge[i].v])
			swap(edge[i].u, edge[i].v);
		val[id[edge[i].u]] = edge[i].w;
	}
	Build(1, n, 1);
	while(q--) {
		scanf("%d", &op);
		if(op == 0) {
			scanf("%d", &a);
			printf("%d\n", GetAns(s, a));
			s = a;
		}
		else {
			scanf("%d %d", &a, &b);
			Update(id[edge[a].u], b, 1, n, 1);
		}
	}
	
	return 0;
}

原文链接:bLue’s Blog

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