20201115gryz模拟赛解题报告
写在前面
T1：期望100pts，实际0pts（7:50 ~ 8:50
T2：期望0pts，实际0pts（10:00 ~ 10:35
T3：期望20pts，实际40pts( 9:10 ~ 10:00，10:35 ~ 11:00
排名：rank2/10
感觉今天状态好多了~ >-<!
今天修树剖太累了，先咕咕咕，有空再把剩下两道题补上还有T3的满分代码，嘻嘻
解题过程
有了CSP的经验，先通读一遍题目，果断开T1，发现数的位数比较小，感觉暴力可以过，但是全排列不会写，发现贪心好像更简单，胡乱搞了两下，过了样例，判了前导零
感觉没问题了,去吃了四楼顿饭，接着开T2，手模样例 + 乱搞，以为是个找规律，好像并不会做，暴力也不会打，
决定先打T3暴力，读过一遍题目，发现这就是个裸树剖，算着暴力会T，为了骗分，暴力搞上再说，此时刚好打9：30下课铃，9:50，上课铃响，稍微一调，过了样例
（我直接感动哭了，树剖从来没有打这么快过，还没有写挂）
继续手模T2，没整明白，后来想T3优化也不成，最后五分钟发现T1忘记判负了，懒得改了~
（后来发现T1没负数，我贪心贪错了
正解
T1
Solution
暴力枚举全排列，复杂度最多 \(10 \times 9!\) , 轻松A掉
Talk is cheap，show me the code：
next_permutation函数不太会用啊，调了一下午树剖，太累了，懒得写了
我的贪心(不忍直视：
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;
int T, k;
string s1, s2;
int read(){
int s = 0, w = 1;
char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') w = -1; ch = getchar(); }
while(ch >= '0' && ch <= '9') s = (s << 1) + (s << 3) + ch - '0', ch = getchar();
return s * w;
}
int main(){
//freopen("cooperate.in","r",stdin);
//freopen("cooperate.out","w",stdout);
T = read();
while(T--){
cin>>s1; k = read();
s2 = s1;
int len = s1.size(), wz, k2 = k;
//最大值
for(int i = 0; i < len; ++i){
wz = i;
for(int j = i + 1; j < len; ++j){
if(s1[i] < s1[j] && s1[wz] < s1[j]){
wz = j;
}
}
if(wz != i && k){
int x = s1[i];
s1[i] = s1[wz]; s1[wz] = x;
k--;
}
}
//最小值
wz = 0;
for(int j = 1; j < len; ++j){//最小值需要考虑不能有前导零，只要保证第一位不是就好了
if(s2[0] > s2[j] && s2[wz] >= s2[j] && s2[j] != '0'){
wz = j;
}
}
if(wz != 0 && k2){
int x = s2[0];
s2[0] = s2[wz], s2[wz] = x;
k2--;
}
for(int i = 1; i < len; ++i){
wz = i;
for(int j = i + 1; j < len; ++j){
if(s2[i] > s2[j] && s2[wz] >= s2[j]){
wz = j;
}
}
if(wz != i && k2){
int x = s2[i];
s2[i] = s2[wz], s2[wz] = x;
k2--;
}
}
int sum1 = 0, sum2 = 0;
for(int i = 0; i < len; ++i){
sum1 = (sum1 << 1) + (sum1 << 3) + s1[i] - '0';
sum2 = (sum2 << 1) + (sum2 << 3) + s2[i] - '0';
}
// cout<<sum1<<endl<<sum2<<endl;
printf("%d\n", sum1 - sum2);
}
return 0;
}
/*
8
12 1
213 2
998244350 1
998244350 2
998244353 1
998244353 2
998244353 3
998244353 300
*/
T2
Solution
把整个区间分成 \(k (或者 k + 1)\) 份，发现翻转两次就能使两盏处在同一位置的灯都点亮，
然后……然后还没看呢
T3
solution
（这好像是今天唯一的成果）
裸的树剖，在求和是注意优化， \(n ^ 2\) 暴力只有20pts
sbw给出了优化方案（O(n)遍历所有边求和，70pts：
\[ans_x = \sum_{j \in son_x} w_j * (siz_x - siz_j) * siz_j \bmod 2019
\]
Talk is cheap，show me the code：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
const int MAXN = 5e4+5;
const int mod = 2019;
int n, Q, cnt;
int dfn[MAXN], pre[MAXN], fath[MAXN], son[MAXN], dep[MAXN], top[MAXN], siz[MAXN], a[MAXN], rd[MAXN];
int read(){
int s = 0, w = 1;
char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') w = -1; ch = getchar(); }
while(ch >= '0' && ch <= '9') s = (s << 1) + (s << 3) + ch - '0', ch = getchar();
return s * w;
}
namespace Seg{
#define lson i << 1
#define rson i << 1 | 1
struct Tree{
int sum, lazy, len;
}tree[MAXN << 2];
void push_up(int i){
tree[i].sum = (tree[lson].sum + tree[rson].sum) % mod;
}
void push_down(int i){
if(tree[i].lazy){
tree[lson].lazy = (tree[lson].lazy + tree[i].lazy) % mod;
tree[rson].lazy = (tree[rson].lazy + tree[i].lazy) % mod;
tree[lson].sum = (tree[lson].sum + tree[i].lazy * tree[lson].len) % mod;
tree[rson].sum = (tree[rson].sum + tree[i].lazy * tree[rson].len) % mod;
tree[i].lazy = 0;
}
}
void build(int i, int l, int r){
tree[i].lazy = 0, tree[i].len = r - l + 1;
if(l == r){
tree[i].sum = a[pre[l]] % mod;
return ;
}
int mid = (l + r) >> 1;
build(lson, l, mid), build(rson, mid + 1, r);
push_up(i);
return ;
}
void add(int i, int l, int r, int L, int R, int k){
if(L <= l && r <= R){
tree[i].sum = (tree[i].sum + tree[i].len * k) % mod;
tree[i].lazy = (tree[i].lazy + k) % mod;
return ;
}
push_down(i);
int mid = (l + r) >> 1;
if(mid >= L) add(lson, l, mid, L, R, k);
if(mid < R) add(rson, mid + 1, r, L, R, k);
push_up(i);
return ;
}
int get_sum(int i, int l, int r, int L, int R){
if(L <= l && r <= R){
return tree[i].sum % mod;
}
push_down(i);
int mid = (l + r) >> 1, ans = 0;
if(mid >= L) ans = (ans + get_sum(lson, l, mid, L, R)) % mod;
if(mid < R) ans = (ans + get_sum(rson, mid + 1, r, L, R)) % mod;
return ans % mod;
}
}
namespace Cut{
struct edge{
int to, w, nxt;
}e[MAXN << 1];
int head[MAXN], num_edge = 0;
void add_edge(int from, int to, int w){
e[++num_edge] = (edge){to, w, head[from]}, head[from] = num_edge;
}
void dfs(int x, int fa){
siz[x] = 1, dep[x] = dep[fa] + 1, fath[x] = fa;
for(int i = head[x]; i; i = e[i].nxt){
int v = e[i].to;
if(v == fa) continue;
dfs(v, x);
siz[x] += siz[v];
if(siz[v] > siz[son[x]]) son[x] = v;
}
}
void dfs2(int x, int tp){
dfn[x] = ++cnt, pre[cnt] = x, top[x] = tp;
if(son[x]) dfs2(son[x], tp);
for(int i = head[x]; i; i = e[i].nxt){
int v = e[i].to;
if(v == fath[x] || v == son[x]) continue;
dfs2(v, v);
}
}
void change(int x, int y, int k){
while(top[x] != top[y]){
if(dep[top[x]] < dep[top[y]]) swap(x, y);
Seg::add(1, 1, n, dfn[top[x]], dfn[x], k);
x = fath[top[x]];
}
if(dfn[x] > dfn[y]) swap(x ,y);
Seg::add(1, 1, n, dfn[x] + 1, dfn[y], k);
return ;
}
int get(int x, int y){
int ans = 0;
while(top[x] != top[y]){
if(dep[top[x]] < dep[top[y]]) swap(x, y);
ans = (ans + Seg::get_sum(1, 1, n, dfn[top[x]], dfn[x])) % mod;
x = fath[top[x]];
}
if(dfn[x] > dfn[y]) swap(x, y);
ans = (ans + Seg::get_sum(1, 1, n, dfn[x] + 1, dfn[y])) % mod;
return ans % mod;
}
int get_s(int x, int rt){//O(n)遍历所有边求和，正解，100pts
int ans = 0;
for(int i = head[x]; i; i = e[i].nxt){
int v = e[i].to;
if(v == fath[x]) continue;
ans = (ans + Seg::get_sum(1, 1, n, dfn[v], dfn[v]) % mod * (siz[rt] - siz[v]) % mod * siz[v]) % mod;
ans = (ans + get_s(v, rt)) % mod;
}
return ans;
}
int get_ss(int x){//暴力，枚举所有情况求和，只有40pts
int ans = 0;
for(int i = dfn[x]; i < dfn[x] + siz[x] - 1; ++i){
for(int j = i + 1; j <= dfn[x] + siz[x] - 1; ++j){
ans = (ans + get(pre[i], pre[j])) % mod;
}
}
return ans;
}
}
int main(){
//freopen("network.in","r",stdin);
//freopen("network.out","w",stdout);
n = read(), Q = read();
for(int i = 2, u; i <= n; ++i){
u = read(), a[i] = read();
Cut::add_edge(u, i, a[i]), Cut::add_edge(i, u, a[i]);
rd[i]++;
}
int wz;
for(int i = 1; i <= n; ++i) if(!rd[i]) {wz = i; break; }
Cut::dfs(wz, 0), Cut::dfs2(wz, wz), Seg::build(1, 1, n);
for(int i = 1, x, y, k; i <= Q; ++i){
string s;
cin>>s;
if(s[0] == 'I'){
x = read(), y = read(), k = read();
Cut::change(x, y, k);
}
if(s[0] == 'A'){
x = read();
// printf("%d\n", Cut::get_ss(x) % mod);
printf("%d\n", Cut::get_s(x, x) % mod);
}
}
return 0;
}
/*
5 5
1 1
2 5
1 2
2 1
INC 2 4 2
INC 3 4 1
ASK 2
INC 2 5 3
ASK 1
*/
