LoveFrieren @ 2024-08-20 19:46:25
刚刚学完就来敲了,敲完还和答案核对了一遍,没任何差别(可能我有的地方没看到?)
代码:
#include<bits/stdc++.h>
using namespace std;
const int M = 1e5 + 10; typedef long long ll;
int n, m, r, p, cnt, wson[M], vl[M], sz[M], fa[M], dfn[M], rdfn[M], top[M], d[M];
ll sgt[M * 4];
vector<int> tr[M];
ll read(){
int f = 1, a = 0;
char c = getchar();
while(c < '0' || c > '9'){
if (c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9'){
a = (a << 1) + (a << 3) + (c ^ 48);
c = getchar();
}
return f * a;
}
dfs1(int u, int f){//求父节点, 重儿子
fa[u] = f;
sz[u] = 1;
for(int i = 0; i < tr[u].size(); ++i){
int v = tr[u][i];
if(v == f) continue;
d[v] = d[u] + 1;
dfs1(v, u);
sz[u] += sz[v];
if(sz[v] > sz[wson[u]]) wson[u] = v;
}
}
dfs2(int u, int topp){//求重链顶,dfs序
dfn[u] = ++cnt;
rdfn[cnt] = u;
top[u] = topp;
if(wson[u] != 0){
dfs2(wson[u], topp);
for(int i = 0; i < tr[u].size(); ++i)
dfs2(tr[u][i], tr[u][i]);
}
}
ll lzt[M * 4];//线段树,因为看过MLE的案例所以没写结构体,见谅
void pushup(int u){sgt[u] = sgt[u * 2] + sgt[u * 2 + 1] % p;}
void build(int u, int L, int R){
if(L == R){
sgt[u] = vl[rdfn[L]];
return;
}
int mid = L + (R - L >> 1);
build(u * 2, L, mid), build(u * 2 + 1, mid + 1, R);
pushup(u);
}
void maketag(int u, int len, int vai){
vl[u] += vai * len % p;
lzt[u] += vai % p;
lzt[u] %= p, vl[u] %= p;
}
void pushdown(int u, int L, int R){
int mid = L + (R - L >> 1);
maketag(u * 2, mid - L + 1, lzt[u]);
maketag(u * 2 + 1, R - mid, lzt[u]);
lzt[u] = 0;
}
void update(int u, int L, int R, int l, int r, int vai){
if(l <= L && R <= r){
maketag(u, R - L + 1, vai);
}
else if(!(r < L || R < l)){
int mid = L + (R - L >> 1);
pushdown(u, L, R);
update(u * 2, L, mid, l, r, vai);
update(u * 2 + 1, mid + 1, R, l, r, vai);
pushup(u);
}
}
ll query(int u, int L, int R, int l, int r){
if(l <= L && R <= r)
return sgt[u];
else if(!(r < L || R < l)){
int mid = L + (R - L >> 1);
pushdown(u, L, R);
return (query(u * 2, L, mid, l, r) + query(u * 2 + 1, mid + 1, R, l, r));
}
else return 0;
}
void upd(int u, int v, int vai){//题目描述上的1,2操作
while(top[u] != top[v]){
if(d[top[u]] < d[top[v]]) swap(u, v);
update(1, 1, n, dfn[top[u]], dfn[u], vai);
u = fa[top[u]];
}
update(1, 1, n, min(dfn[u], dfn[v]), max(dfn[u], dfn[v]), vai);
}
ll qry(int u, int v){
ll re = 0;
while(top[u] != top[v]){
if(d[top[u]] < d[top[v]]) swap(u, v);
re += query(1, 1, n, dfn[top[u]], dfn[u]) % p;
re %= p;
u = fa[top[u]];
}
re += query(1, 1, n, min(dfn[u], dfn[v]), max(dfn[u], dfn[v])) % p;
return re % p;
}
int main(){
n = read(), m = read(), r = read(), p = read();
for(int i = 1; i <= n; ++i) vl[i] = read();
for(int i = 1, u, v; i < n; ++i){
u = read(), v = read();
tr[u].push_back(v);
tr[v].push_back(u);
}
build(1, 1, n); ++d[r]; dfs1(r, 0); dfs2(r, 0);
for(int i = 1, x, y, z, t; i <= m; ++i){
t = read();
if(t == 1){
x = read(), y = read(), z = read();
upd(x, y, z);
}
else if(t == 2){
x = read(), y = read();
cout << qry(x, y) << "\n";
}
else if (t == 3){
x = read(), y = read();
update(1, 1, n, dfn[x], dfn[x] + sz[x] - 1, y);
}
else{
x = read();
cout << query(1, 1, n, dfn[x], dfn[x] + sz[x] - 1) % p << "\n";
}
}
return 0;
}
by Grammar__hbw @ 2024-08-20 20:11:41
@ckkjz 树剖的时候传的区间有时候会+1
by Grammar__hbw @ 2024-08-20 20:13:22
@ckkjz dfs2错了,遍历树的时候要判断不等于父亲及重儿子,但你没有
by Grammar__hbw @ 2024-08-20 20:13:48
@ckkjz val * len会爆int
by LoveFrieren @ 2024-08-20 20:17:19
@Grammar__hbw 就是这个,ok了!(单纯是有输出了,对不对另说())
by LoveFrieren @ 2024-08-20 20:19:50
@Grammar__hbw 为什么有的时候会+1,不是从该节点到链顶吗,他们的dfs序不是连着的吗,为什么会有时候+1
by Grammar__hbw @ 2024-08-20 20:20:48
@ckkjz 改完应该就可以AC了qwqwq
取模不规范见祖宗
不开long long见祖宗
不判l>r见祖宗
dfs不规范见祖宗
by Grammar__hbw @ 2024-08-20 20:21:41
@ckkjz 有些题目lca不能算(主要是把边权转点权的题目)
by Grammar__hbw @ 2024-08-20 20:23:59
@ckkjz 重剖的dfs2的第二个参数应该是r,否则0作链顶见祖宗qwqwqwqwq
by Grammar__hbw @ 2024-08-20 20:24:55
@ckkjz 把我的线段树modify给你,你自己看看你少了什么
void modify(int l,int r,int ll,int rr,int o,long long v){
if(l<=ll&&rr<=r){val[o]=(val[o]+1ll*v*(rr-ll+1))%mod;lz[o]+=v;return;}
pushdown(ll,rr,o);
int mid=(ll+rr)>>1;
if(l<=mid) modify(l,r,ll,mid,o*2,v);
if(r>mid) modify(l,r,mid+1,rr,o*2+1,v);
update(o);
}
by Grammar__hbw @ 2024-08-20 20:36:15
@ckkjz 重剖的dfs2的第二个参数应该是r,否则0作链顶见祖宗qwqwqwqwq