「codeforces - 1486F」Pairs of Paths

link。

还算萌，但是代码有些难写……

你首先会想要

int n, m, fa[20][300100], pa[300100], dep[300100], cnt[900100];
int ldf[300100], rdf[300100], dfc, qwq;
vi<int> G[300100];
struct path {
    int x, y, sx, sy, lca;
};
void dfs(int x, int pre) {
    fa[0][x] = pa[x] = pre;
    dep[x] = dep[pre]+1;
    ldf[x] = ++dfc;
    for (int i=1; i<20; ++i) {
        fa[i][x] = fa[i-1][fa[i-1][x]];
    }
    for (auto y : G[x]) {
        if (y != pre) {
            dfs(y, x);
        }
    }
    rdf[x] = dfc;
}
int lca(int x, int y) {
    if (dep[x] < dep[y]) {
        swap(x, y);
    }
    for (int i=19; i>=0; --i) {
        if (dep[fa[i][x]] >= dep[y]) {
            x = fa[i][x];
        }
    }
    if (x == y) {
        return x;
    }
    for (int i=19; i>=0; --i) {
        if (fa[i][x] != fa[i][y]) {
            x = fa[i][x], y = fa[i][y];
        }
    }
    return pa[x];
}
int jump(int x, int d) {
    if (d < 0) {
        return n+(++qwq);
    }
    for (int i=19; i>=0; --i) {
        if ((d>>i)&1) {
            x = fa[i][x];
        }
    }
    return x;
}
int bit[300100];
void add(int x, int y) {
    for (; x<=n; x+=x&-x) {
        bit[x] += y;
    }
}
int qry(int x) {
    int res = 0;
    for (; x; x-=x&-x) {
        res += bit[x];
    }
    return res;
}
int qry(int l, int r) {
    return qry(r)-qry(l-1);
}
void executer() {
    cin >> n;
    for (int i=1,x,y; i<n; ++i) {
        cin >> x >> y;
        G[x].push_back(y);
        G[y].push_back(x);
    }
    dfs(1, 0);
    cin >> m;
    vi<path> paths(m);
    for (auto& it : paths) {
        cin >> it.x >> it.y;
        it.lca = lca(it.x, it.y);
        it.sx = jump(it.x, dep[it.x]-dep[it.lca]-1);
        it.sy = jump(it.y, dep[it.y]-dep[it.lca]-1);
        if (it.sx > it.sy) {
            swap(it.sx, it.sy), swap(it.x, it.y);
        }
    }
    sort(paths.begin(), paths.end(), [&](const path& lhs, const path& rhs) {
        return dep[lhs.lca] < dep[rhs.lca]
               || (dep[lhs.lca] == dep[rhs.lca] && lhs.lca < rhs.lca)
               || (lhs.lca == rhs.lca && lhs.sx > rhs.sx)
               || (lhs.sx == rhs.sx && lhs.sy > rhs.sy);
    });
    LL ans = 0;
    for (int i=0; i<m;) {
        int j = i;
        while (j < m-1 && paths[j+1].lca == paths[j].lca) {
            j++;
        }
        LL now = 0;
        for (int l=i; l<=j;) {
            int r = l;
            while (r < j && paths[r+1].sx == paths[r].sx) {
                r++;
            }
            for (int k=l; k<=r; ++k) {
                ans += now-cnt[paths[k].sy];
            }
            for (int k=l; k<=r; ++k) {
                cnt[paths[k].sx]++, cnt[paths[k].sy]++;
            }
            now += r-l+1, l = r+1;
        }
        for (int k=i; k<=j; ++k) {
            cnt[paths[k].sx] = cnt[paths[k].sy] = 0;
        }
        for (int k=i; k<=j; ++k) {
            ans += qry(ldf[paths[k].lca], rdf[paths[k].lca]);
            if (paths[k].sx <= n) {
                ans -= qry(ldf[paths[k].sx], rdf[paths[k].sx]);
            }
            if (paths[k].sy <= n) {
                ans -= qry(ldf[paths[k].sy], rdf[paths[k].sy]);
            }
        }
        for (int k=i; k<=j; ++k) {
            add(ldf[paths[k].x], 1), add(ldf[paths[k].y], 1);
        }
        i = j+1;
    }
    cout << ans << "\n";
}