异象石

题目描述

思路

一本通的描述比较详细
比较好的博客

代码

#include <cstdio>
#include <cstring>
#include <set>
#include <iterator>
#define FOR(a, n) for(int i = 1; i <= (n); ++i)
using namespace std;


const int MAX = 1e5 + 5;
int n, m;
long long ans;
int head[MAX], ver[MAX << 1], edge[MAX << 1], nt[MAX << 1], ht; // 存储边相关
int dfn[MAX], cnt; // dfn序相关 
int f[MAX][21], dep[MAX]; // lca相关
long long dist[MAX];
set<pair<int, int> > st;
set<pair<int, int> >::iterator it;
void add(int x, int y, int z) {
	nt[++ht] = head[x], head[x] = ht, ver[ht] = y, edge[ht] = z;
}

void dfs_lca(int x, int u, int z) {
	dep[x] = dep[u] + 1;
	dfn[x] = ++cnt;
	dist[x] = dist[u] + z;
	f[x][0] = u;
	for (int i = 1; i < 21; ++i) {
		f[x][i] = f[f[x][i - 1]][i - 1];
	}
	for (int i = head[x], j, k; i; i = nt[i]) {
		j = ver[i], k = edge[i];
		if (j == u) continue;
		dfs_lca(j, x, k);
	}
}

int lca(int x, int y) {
	int X = x, Y = y;
	if (dep[x] < dep[y]) swap(x, y);
	for (int i = 20; i >= 0; --i) {
		if (dep[f[x][i]] >= dep[y]) {
			x = f[x][i];
		}
	}
	if (x == y) return x;
	for (int i = 20; i >= 0; --i) {
		if (f[x][i] != f[y][i]) {
			x = f[x][i], y = f[y][i];
		}
	}
	// printf("lca: %d %d %d
", X, Y, f[x][0]);
	return f[x][0];
}

long long path(int x, int y) {
	return dist[x] + dist[y] - (dist[lca(x, y)] << 1);
}

inline int read() {
	int s = 0;
	char ch = getchar();
	while (ch < '0' || ch > '9') ch = getchar();
	while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
	return s;
}

void show() {
	printf("%d %d
", n, m);
	puts("dep:");
	FOR(dep, n) printf("%d ", dep[i]); puts("");
	puts("dfn");
	FOR(dfn, n) printf("%d ", dfn[i]); puts("");
	puts("dist");
	FOR(dist, n) printf("%lld ", dist[i]); puts("");
	puts("fa");
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j < 4; j++) {
			printf("%d ", f[i][j]);
		}
		puts("");
	}
}

int main() {
	n = read();
	for (int i = 1, a, b, c; i < n; ++i) {
		a = read(), b = read(), c = read();
		add(a, b, c), add(b, a, c);
	}
	m = read();
	dfs_lca(1, 0, 0);
	// show();
	char ch[5];
	for (int i = 1, j; i <= m; ++i) {
		scanf("%s", ch);
		// puts(ch);
		pair<int, int> l, r, te;
		if (strcmp(ch, "+") == 0) {
			j = read();
			te = make_pair(dfn[j], j);
			if (st.size() != 0) { // 从第二个点开始计算ans的值
				it = (st.lower_bound(te));
				if (it == st.begin()) it = st.end();
				l = *(--it);
				// printf("L +: %d %d
", l.first, l.second);
				it = st.upper_bound(te);
				if (it == st.end()) r = *st.begin();
				else r = *it;
				// printf("R +: %d %d
", r.first, r.second);
				ans = ans - path(l.second, r.second) + path(l.second, j) + path(r.second, j);
			}
			st.insert(make_pair(dfn[j], j)); // 第一个点直接插入
		} else if (strcmp(ch, "-") == 0) {
			j = read();
			te = make_pair(dfn[j], j);
			st.erase(te);
			if (st.size() == 0) {  // 删除最后一个点之后，直接返回
				ans = 0;
				continue;
			}
			it = st.lower_bound(te); // 剩余点超过0个，要重新计算ans的值
			if (it == st.begin()) it = st.end();
			l = *(--it);
			// printf("L -: %d %d
", l.first, l.second);
			it = st.upper_bound(te);
			if (it == st.end()) r = *st.begin();
			else r = *it;
			// printf("R -: %d %d
", r.first, r.second);
			ans = ans + path(l.second, r.second) - path(l.second, j) - path(r.second, j);
		} else {
			printf("%lld
", ans / 2);
		}
	}
	return 0;
}