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\n", 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", 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) {
                it = (st.lower_bound(te));
                if (it == st.begin()) it = st.end();
                l = *(--it);
                // printf("L +: %d %d\n", l.first, l.second);
                it = st.upper_bound(te);
                if (it == st.end()) r = *st.begin();
                else r = *it;
                // printf("R +: %d %d\n", 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));
        } else if (strcmp(ch, "-") == 0) {
            j = read();
            te = make_pair(dfn[j], j);
            st.erase(te);
            if (st.size() == 0) {
                ans = 0;
                continue;
            }
            it = st.lower_bound(te);
            if (it == st.begin()) it = st.end();
            l = *(--it);
            // printf("L -: %d %d\n", l.first, l.second);
            it = st.upper_bound(te);
            if (it == st.end()) r = *st.begin();
            else r = *it;
            // printf("R -: %d %d\n", r.first, r.second);
            ans = ans + path(l.second, r.second) - path(l.second, j) - path(r.second, j);
        } else {
            printf("%lld\n", ans / 2);
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/liuzz-20180701/p/11505157.html