来自学长的馈赠6
A. ^_^
考虑一个节点有贡献，当且仅当它比子树内所有点先被选中，那么显然答案就是
\(\sum1/size[x]\)
code
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
inline int read(){
int x = 0; char c = getchar();
while(c < '0' || c > '9')c = getchar();
do{x = (x << 3) + (x << 1) + (c ^ 48); c = getchar();}while(c >= '0' && c <= '9');
return x;
}
typedef long long ll;
const int maxn = 10000005;
const ll mod = 998244353;
ll qpow(ll x, int y){
ll ans = 1;
for(;y;y>>=1,x =x * x % mod)if(y & 1)ans = ans * x % mod;
return ans;
}
int size[maxn], fa[maxn];
ll inv[maxn];
int main(){
int n = read();
for(int i = 2; i <= n; ++i)fa[i] = read();
for(int i = n; i >= 1; --i){
++size[i];
size[fa[i]] += size[i];
}
for(ll i = 1; i <= n; ++i)inv[i] = qpow(i, mod - 2);
ll ans = 0;
for(int i = 1; i <= n; ++i){
ans = (ans + inv[size[i]]) % mod;
}
printf("%lld\n",ans);
return 0;
}
B. 软件包管理器
树剖板子
树剖维护的\(dfs\)序能够处理子树内问题！！！！！
code
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
inline int read(){
int x = 0; char c = getchar();
while(c < '0' || c > '9')c = getchar();
do{x = (x << 3) + (x << 1) + (c ^ 48); c = getchar();}while(c >= '0' && c <= '9');
return x;
}
typedef long long ll;
const int maxn = 100005;
int n, fa[maxn], q, tot, head[maxn];
char c[15];
int son[maxn], top[maxn], dep[maxn], size[maxn], id[maxn];
struct edge{
int to,net;
}e[maxn << 1 | 1];
void add(int u, int v){
e[++tot].net = head[u];
head[u] = tot;
e[tot].to = v;
}
void dfs1(int x){
size[x] = 1;
for(int i = head[x]; i; i = e[i].net){
int v = e[i].to;
fa[v] = x; dep[v] = dep[x] + 1;
dfs1(v);
size[x] += size[v];
if(size[son[x]] < size[v])son[x] = v;
}
}
int tim;
void dfs2(int x, int tp){
top[x] = tp; id[x] = ++tim;
if(son[x])dfs2(son[x], tp);
for(int i = head[x]; i; i = e[i].net){
int v = e[i].to;
if(v == son[x])continue;
dfs2(v, v);
}
}
struct tree{
struct node{
int size,tag;
}t[maxn << 2 | 1];
void push_up(int x){t[x].size = t[x << 1].size + t[x << 1 | 1].size;}
void push_down(int x, int l, int r){
int ls = x << 1,rs = x << 1 | 1, mid = (l + r) >> 1;
t[ls].tag = t[rs].tag = t[x].tag;
t[ls].size = t[rs].size = 0;
if(t[x].tag == 1){
t[ls].size = mid - l + 1;
t[rs].size = r - mid;
}
t[x].tag = 0;
}
void modify(int x, int l, int r, int L, int R, int dt){
if(L <= l && r <= R){
t[x].size = dt * (r - l + 1);
t[x].tag = dt ? 1 : -1;
return;
}
if(t[x].tag)push_down(x, l, r);
int mid = (l + r) >> 1;
if(L <= mid)modify(x << 1, l, mid, L, R, dt);
if(R > mid)modify(x << 1 | 1, mid + 1, r, L, R, dt);
push_up(x);
}
int query(int x, int l, int r, int L, int R){
if(L <= l && r <= R)return t[x].size;
if(t[x].tag)push_down(x, l, r);
int mid = (l + r) >> 1, ans = 0;
if(L <= mid)ans += query(x << 1, l, mid, L, R);
if(R > mid)ans += query(x << 1 | 1, mid + 1, r, L, R);
return ans;
}
}t;
void install(int x){
int u = x, cnt = 0;
while(u){
cnt += t.query(1, 1, n, id[top[u]], id[u]);
t.modify(1, 1, n, id[top[u]], id[u], 1);
u = fa[top[u]];
}
cnt = dep[x] - cnt;
printf("%d\n",cnt);
}
void uninstall(int x){
int cnt = t.query(1, 1, n, id[x], id[x] + size[x] - 1);
t.modify(1, 1, n, id[x], id[x] + size[x] - 1, 0);
printf("%d\n",cnt);
}
int main(){
n = read();
for(int i = 2; i <= n; ++i){
fa[i] = read() + 1;
add(fa[i], i);
}
dep[1] = 1; fa[1] = 0;
dfs1(1); dfs2(1,1);
q = read();
for(int i = 1; i <= q; ++i){
scanf("%s",c);
int x = read() + 1;
if(c[0] == 'i')install(x);
else uninstall(x);
}
return 0;
}
C. 地理课
对每条边找出其作用的区间，然后在线段树上修改对应区间
进行一次线段树\(DFS\)，配合可以撤销的并查集，按秩合并
code
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <map>
#include <queue>
#include <vector>
using namespace std;
inline int read() {
int x = 0;char c = getchar();
while (c < '0' || c > '9') c = getchar();
do {x = (x << 3) + (x << 1) + (c ^ 48);c = getchar();
} while (c >= '0' && c <= '9');
return x;
}
typedef long long ll;
const int maxn = 200005;
const int mod = 1e9 + 7;
int qpow(int x, int y) {
int ans = 1;
for (; y; y >>= 1, x = 1ll * x * x % mod)
if (y & 1) ans = 1ll * ans * x % mod;
return ans;
}
int n, m, top;
ll inv[maxn];
struct stac{
int s, f1, f2, id;
}rem[maxn << 3 | 1];
struct SET{
int f[maxn], size[maxn];
void pre() {
for (int i = 1; i <= n; ++i) f[i] = i;
for (int i = 1; i <= n; ++i) size[i] = 1;
}
int fa(int x) {
return f[x] == x ? x : fa(f[x]);
}
} s;
struct op {
int op, x, y, r;
} o[maxn];
int ans[maxn];
struct Tree{
vector<int> v[maxn << 2 | 1];
void modify(int x, int l, int r, int L, int R, int now) {
if (L <= l && r <= R) {
v[x].push_back(now);
return;
}
int mid = (l + r) >> 1;
if (L <= mid) modify(x << 1, l, mid, L, R, now);
if (R > mid) modify(x << 1 | 1, mid + 1, r, L, R, now);
}
ll nans = 1;
void LINK(int x) {
for (int i : v[x]) {
int u = o[i].x, v = o[i].y;
u = s.fa(u);
v = s.fa(v);
if (u == v) continue;
if (s.size[u] < s.size[v]) swap(u, v);
nans = nans * inv[s.size[u]] % mod * inv[s.size[v]] % mod;
rem[++top].f1 = v;
rem[top].f2 = u;
rem[top].s = s.size[v];
rem[top].id = x;
s.size[u] += s.size[v];
s.size[v] = 0;
nans = nans * s.size[u] % mod;
s.f[v] = u;
}
}
void CUT(int x) {
while(rem[top].id == x){
stac &o = rem[top];
if(o.id != x)return;
s.f[o.f1] = o.f1;
s.f[o.f2] = o.f2;
nans = nans * inv[s.size[o.f2]] % mod;
s.size[o.f2] -= o.s;
s.size[o.f1] = o.s;
nans = nans * s.size[o.f1] % mod * s.size[o.f2] % mod;
--top;
}
}
void work(int x, int l, int r) {
// printf("%d %lld\n",x,nans);
LINK(x);
if (l == r) {
ans[l] = nans;
CUT(x);
return;
}
int mid = (l + r) >> 1;
work(x << 1, l, mid);
work(x << 1 | 1, mid + 1, r);
CUT(x);
}
} t;
typedef pair<int, int> pii;
map<pii, int> mp;
int main() {
n = read(), m = read();
inv[1] = 1;
for (int i = 2; i <= n; i++)
inv[i] = (mod - mod / i) * inv[mod % i] % mod;
s.pre();
for (int i = 1; i <= m; ++i) {
o[i].op = read();
o[i].x = read();
o[i].y = read();
if (o[i].x > o[i].y) swap(o[i].x, o[i].y);
if (o[i].op == 2) {
o[mp[pii(o[i].x, o[i].y)]].r = i - 1;
}else mp[pii(o[i].x, o[i].y)] = i;
}
for (int i = 1; i <= m; ++i)
{
if (o[i].op == 1) t.modify(1, 1, m, i, o[i].r ? o[i].r : m, i);
}
t.work(1, 1, m);
for (int i = 1; i <= m; ++i)printf("%d\n",ans[i]);
return 0;
}
D. 道路和航线
考场写出数据生成器，完美展现了题解思路
然而就是没写出正解，乱改\(dij\)还把\(spfa\)的\(30\)分改没了
正解就是对不同联通块整体拓扑，每次把一个块内点全部入队
rand
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
mt19937 rd((ull)(new char) * (ull) (new char));
int sr(int l, int r){
uniform_int_distribution<>d(l, r);
return d(rd);
}
int bl[25005];
int lim = 10;
int main(){
int t = 5, r = 3, p = 3;
int s = sr(1, t);
int fk = 20;
for(int i = 1; i <= t; ++i){
bl[i] = sr(1,fk);//(i % fk) + 1;
}
printf("%d %d %d %d\n",t,r,p,s);
for(int i = 1; i <= r; ++i){
int u = sr(1, t), v = sr(1, t);
while(bl[u]!=bl[v]){
u = sr(1, t), v = sr(1, t);
}
printf("%d %d %d\n",u,v,sr(0,lim));
}
for(int i = 1; i <= p; ++i){
int u = sr(1, t);
int v = sr(1, t);
while(bl[u] >= bl[v]){
u = sr(1, t);
v = sr(1,t);
}
printf("%d %d %d\n",u,v,sr(-lim,lim));
}
return 0;
}
code
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
inline int read(){
int x = 0; bool f = 0; char c = getchar();
while(c < '0' || c > '9'){if(c == '-')f = 1;c = getchar();}
do{x = (x << 3) + (x << 1) + (c ^ 48); c = getchar();}while(c >= '0' && c <= '9');
return f ? -x: x;
}
typedef long long ll;
typedef pair<ll, int> pli;
const int maxn = 250055;
struct edge{
int net,to,val;
}e[maxn << 1 | 1];
int head[maxn], tot;
ll dis[maxn];
void add(int u, int v, int val){
e[++tot].net = head[u];
head[u] = tot;
e[tot].to = v;
e[tot].val = val;
}
int t,r,p,s;
bool vis[maxn];
int belong[maxn], tim, rd[maxn];
vector<int> node[maxn];
void dfs(int x){
belong[x] = tim;
node[tim].push_back(x);
for(int i = head[x]; i; i = e[i].net){
int v = e[i].to;
if(belong[v])continue;
dfs(v);
}
}
priority_queue<pli, vector<pli>, greater<pli> >q;
int main(){
t = read(), r = read(), p = read(), s = read();
for(int i = 1; i <= r; ++i){
int a, b, c; a = read(), b = read(), c = read();
add(a, b, c);add(b, a, c);
}
for(int i = 1; i <= t; ++i)if(!belong[i])++tim, dfs(i);
for(int i = 1; i <= p; ++i){
int a, b, c; a = read(), b = read(), c = read();
add(a, b, c); ++rd[belong[b]];
}
memset(dis,0x3f,sizeof(dis));
dis[s] = 0;
for(int i = 1; i <= tim; ++i)
if(rd[i] <= 0) for(int j : node[i])q.push(make_pair(dis[j],j));
while(!q.empty()){
int x = q.top().second; q.pop();
if(vis[x])continue;
vis[x] = 1;
for(int i = head[x]; i; i = e[i].net){
int v = e[i].to;
if(belong[v] != belong[x]){
--rd[belong[v]];
if(rd[belong[v]] <= 0) for(int y : node[belong[v]])q.push(make_pair(dis[y], y));
}
if(dis[x] == dis[maxn - 1])continue;
if(dis[v] > dis[x] + e[i].val){
dis[v] = dis[x] + e[i].val;
if(rd[belong[v]] <= 0){
q.push(make_pair(dis[v], v));
}
}
}
}
for(int i = 1; i <= t; ++i)
if(dis[i] == dis[maxn - 1])printf("NO PATH\n");
else printf("%lld\n",dis[i]);
return 0;
}
