标签：style http os io ar for 代码 sp amp
题目链接：点击打开链接
题意：
rt。。
在询问时，两端向上爬时记录从深度浅的到深度深的方向上的 （也就是左最大连续子段和）
最后两个点在同一条重链上时合并。
合并时要注意有4种情况， 详见代码。
线段树部分和5相似。
#include <cstdio>
#include <iostream>
#include <string.h>
#include <vector>
using namespace std;
inline void rd(int &n){
n = 0;
bool tmp = 0;
char c = getchar();
while((c < '0' || c > '9') && c != '-' ) c = getchar();
if(c == '-') c = getchar(), tmp = 1;
while(c >= '0' && c <= '9') n *= 10, n += (c - '0'),c = getchar();
if(tmp) n = -n;
}
#define N 100100
#define hehe 10001
#define L(x) tree[x].l
#define R(x) tree[x].r
#define Lson(x) (x<<1)
#define Rson(x) (x<<1|1)
#define Sum(x) tree[x].sum
#define Max(x) tree[x].max
#define Lmax(x) tree[x].lmax
#define Rmax(x) tree[x].rmax
#define Lazy(x) tree[x].lazy
#define Siz(x) tree[x].siz()
struct node{
int l, r;
int mid(){return (l+r)>>1;}
int siz(){return r-l+1;}
int lmax, rmax, max, sum, lazy;
}tree[N<<2];
int n, a[N], Q, fw[N];
void updata_point(int val, int id){Lmax(id) = Rmax(id) = Max(id) = Sum(id) = val * Siz(id);}
void push_down(int id){
if(L(id) == R(id))return ;
if(Lazy(id)!=hehe) {
updata_point(Lazy(id), Lson(id));
updata_point(Lazy(id), Rson(id));
Lazy(Lson(id)) = Lazy(Rson(id)) = Lazy(id);
Lazy(id) = hehe;
}
}
void push_up(int id){
if(L(id)==R(id))return ;
Lmax(id) = max(Lmax(Lson(id)), Sum(Lson(id)) + Lmax(Rson(id)));
Rmax(id) = max(Rmax(Rson(id)), Sum(Rson(id)) + Rmax(Lson(id)));
Sum(id) = Sum(Lson(id)) + Sum(Rson(id));
Max(id) = max(max(Max(Lson(id)), Max(Rson(id))), Rmax(Lson(id)) + Lmax(Rson(id)));
}
void build(int l, int r, int id){
L(id) = l; R(id) = r;
Lazy(id) = hehe;
if(l == r)
{
updata_point(a[fw[l]], id);
return;
}
int mid = tree[id].mid();
build(l, mid, Lson(id));
build(mid+1, r, Rson(id));
push_up(id);
}
void updata(int l, int r, int val, int id){
push_down(id);
if(l == L(id) && R(id) == r)
{
Lazy(id) = val;
updata_point(val, id);
return ;
}
int mid = tree[id].mid();
if(mid < l)
updata(l, r, val, Rson(id));
else if(r <= mid)
updata(l, r, val, Lson(id));
else {
updata(l, mid, val, Lson(id));
updata(mid+1, r, val, Rson(id));
}
push_up(id);
}
int query_sum(int l, int r, int id){
push_down(id);
if(l == L(id) && R(id) == r) return Sum(id);
int mid = tree[id].mid();
if(mid < l)
return query_sum(l, r, Rson(id));
else if(r <= mid)
return query_sum(l, r, Lson(id));
else
return query_sum(l, mid, Lson(id)) + query_sum(mid+1, r, Rson(id));
}
int query_L(int l, int r, int id){
if(l == L(id) && R(id) == r) return Lmax(id);
int mid = tree[id].mid();
if(mid < l)
return query_L(l, r, Rson(id));
else if(r <= mid)
return query_L(l, r, Lson(id));
int lans = query_L(l, mid, Lson(id)), rans = query_L(mid+1, r, Rson(id));
return max(lans, query_sum(l, mid, id) + rans);
}
int query_R(int l, int r, int id){
if(l == L(id) && R(id) == r) return Rmax(id);
int mid = tree[id].mid();
if(mid < l)
return query_R(l, r, Rson(id));
else if(r <= mid)
return query_R(l, r, Lson(id));
int lans = query_R(l, mid, Lson(id)), rans = query_R(mid+1, r, Rson(id));
return max(rans, query_sum(mid+1, r, id) + lans);
}
int query_l(int l, int r, int id){
push_down(id);
if(l == L(id) && R(id) == r) return Lmax(id);
int mid = tree[id].mid();
if(mid < l)
return query_l(l, r, Rson(id));
else if(r <= mid)
return query_l(l, r, Lson(id));
int lans = query_l(l, mid, Lson(id)), rans = query_l(mid+1, r, Rson(id));
return max(lans, Sum(Lson(id)) + rans);
}
int query_r(int l, int r, int id){
push_down(id);
if(l == L(id) && R(id) == r) return Rmax(id);
int mid = tree[id].mid();
if(mid < l)
return query_r(l, r, Rson(id));
else if(r <= mid)
return query_r(l, r, Lson(id));
int lans = query_r(l, mid, Lson(id)), rans = query_r(mid+1, r, Rson(id));
return max(rans, Sum(Rson(id)) + lans);
}
int query(int l, int r, int id){
push_down(id);
if(l == L(id) && R(id) == r)return Max(id);
int mid = tree[id].mid();
if(mid < l)
return query(l, r, Rson(id));
else if(r<=mid)
return query(l, r, Lson(id));
int lans = query(l, mid, Lson(id)), rans = query(mid+1, r, Rson(id));
int ans = max(lans, rans);
return max(ans, query_r(l, mid, Lson(id)) + query_l(mid+1, r, Rson(id)));
}
vector<int>G[N];
int fa[N], son[N], siz[N], dep[N], tree_id;
//父节点 重儿子 子树节点数 深度 线段树标号
int w[N], p[N];
//父边在线段树中的标号 节点顶端的点
void dfs(int u, int Father, int deep){
fa[u] = Father; son[u] = 0; dep[u] = deep; siz[u] = 1;
for(int i = 0; i < G[u].size(); i++) {
int v = G[u][i]; if(v == Father) continue;
dfs(v, u, deep+1);
siz[u] += siz[v];
if(siz[v] > siz[son[u]])son[u] = v;
}
}
void Have_p(int u, int Father){
w[u] = ++ tree_id; fw[tree_id] = u; p[u] = Father;
if(son[u])
Have_p(son[u], Father);
else return ;
for(int i = 0; i < G[u].size(); i++) {
int v = G[u][i]; if(v == fa[u] || v == son[u])continue;
Have_p(v, v);
}
}
void Updata(int l, int r, int val){
int f1 = p[l], f2 = p[r];
while(f1 != f2) {
if(dep[f1] < dep[f2])
swap(f1, f2), swap(l, r);
updata(w[f1], w[l], val, 1);
l = fa[f1]; f1 = p[l];
}
if(dep[l] > dep[r]) swap(l, r);
updata(w[l], w[r], val, 1);
}
int Query(int l, int r){
int ans = 0;
int f1 = p[l], f2 = p[r], lmax = 0, rmax = 0;
while(f1 != f2) {
if(dep[f1] < dep[f2])
swap(f1, f2), swap(l, r), swap(lmax, rmax);
ans = max(ans, query(w[f1], w[l], 1));
ans = max(ans, max(0,query_R(w[f1], w[l], 1)) + lmax);
lmax = max(query_sum(w[f1], w[l], 1) + lmax, query_L(w[f1], w[l], 1));
lmax = max(0, lmax);
l = fa[f1]; f1 = p[l];
}
if(dep[l] > dep[r]) swap(l, r), swap(lmax, rmax);
ans = max(ans, query(w[l], w[r], 1));
ans = max(ans, query_L(w[l], w[r], 1) + lmax);
ans = max(ans, query_R(w[l], w[r], 1) + rmax);
ans = max(ans, query_sum(w[l], w[r], 1) + lmax + rmax);
return ans;
}
void solve(){
siz[0] = tree_id = 0;
dfs(1, 1, 1);
Have_p(1, 1);
build(1, n, 1);
rd(Q); int op, a, b, c;
while(Q--)
{
rd(op); rd(a); rd(b);
if(op == 1)
printf("%d\n", Query(a, b));
else
{
rd(c);
Updata(a, b, c);
}
}
}
void input(){
for(int i = 1; i <= n; i++)rd(a[i]), G[i].clear();
for(int u, v, i = 1; i < n; i++) {
rd(u); rd(v);
G[u].push_back(v); G[v].push_back(u);
}
}
int main(){
while(~scanf("%d",&n)){
input();
solve();
}
return 0;
}
/*
7
-5 7 4 3 -5 8 -3
1 2
1 3
2 4
2 5
5 6
5 7
8
1 4 3
2 1 1 -3
1 3 4
1 2 6
1 6 2
1 2 7
1 7 5
1 4 7
ans :
10
11
10
10
7
0
10
*/
Spoj 6779 Can you answer these queries VII 树链剖分 在树上任意路径的最大子段和 区间修改点权
标签：style http os io ar for 代码 sp amp
原文地址：http://blog.csdn.net/qq574857122/article/details/39004831
