第一题：找LCA，两点之间的距离=他们各自到起点的距离 - 2*LCA到起点的距离
#include<bits/stdc++.h>
using namespace std;
const int maxn = 100015, P = 20;
int head[2 * maxn],to[2 * maxn],last[2 *maxn],co[2 * maxn],dep[maxn], idx, anc[maxn][P+1],dis[maxn];
void dfs(int u,int from){
//in[u] = ++idx;
anc[u][0] = from;
for(int p = 1; p <= P; p++)
anc[u][p] = anc[anc[u][p-1]][p-1];
for(int i = head[u]; i; i = last[i]){
if(to[i] != from){
dep[to[i]] = dep[u] + 1;
dis[to[i]] = dis[u] + co[i];
dfs(to[i], u);
}
}
}
void adde(int u,int v,int w){
idx++;
last[idx] = head[u];
to[idx] = v;
co[idx] = w;
head[u] = idx;
}
int lca(int u,int v){
if(dep[u] < dep[v])swap(u, v);
int t = dep[u] - dep[v];
for(int p = 0; t; t >>= 1, p++)
if(t & 1)u = anc[u][p];
if(u == v)return v;
for(int p = P; p >=0; p--)
if(anc[u][p] != anc[v][p])
u = anc[u][p], v = anc[v][p];
return anc[u][0];
}
int main(){
freopen ( "distance.in", "r", stdin ) ;
freopen ( "distance.out", "w", stdout ) ;
int n,m;
cin>>n;
for(int i=1;i<n;i++){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
adde( u, v, w );
adde( v, u, w );
}
dfs(1,1);
cin>>m;
for(int i=1;i<=m;i++){
int u ,v;
scanf("%d%d",&u,&v);
int q = lca(u,v);
printf("%d\n",dis[v]+dis[u]-dis[q]*2);
}
}
第二题：先dfs的dfn序，再投影得到一个序列，建线段树
#include<bits/stdc++.h>
using namespace std;
const int oo = 0x3f3f3f3f;
#define MAX_N 100005
long long sum ;
int head[MAX_N<<1],next[MAX_N<<1],to[MAX_N<<1];
int tot, a[MAX_N], in[MAX_N], out[MAX_N],dep[MAX_N],n, m, idx;
void dfs(int u,int from){
in[u] = ++idx;
for(int i = head[u]; i; i = next[i]){
int v = to[i];
if(v != from){
dep[v] = dep[u] + 1;
dfs(v,u);
}
}
out[u] = idx;
}
void adde(int u,int v){
++tot;
next[tot] = head[u];
to[tot] = v;
head[u] = tot;
}
struct SegTree{
struct node{
long long sum;
int vmin,vmax,lazy;
};
node Tree[MAX_N << 2];
#define ls l, m, v << 1
#define rs m+1, r, v << 1 | 1
void push_up(int v){
Tree[v].sum = Tree[v << 1].sum + Tree[v << 1 | 1].sum;
Tree[v].vmin = min(Tree[v << 1].vmin, Tree[v << 1 | 1].vmin);
Tree[v].vmax = max(Tree[v << 1].vmax, Tree[v << 1 | 1].vmax);
}
void push_down(int l, int r, int v){
int m = (l + r) >> 1;
Tree[v << 1].sum += 1LL * Tree[v].lazy * (m - l +1);
Tree[v << 1].vmin += Tree[v].lazy;
Tree[v << 1].vmax += Tree[v].lazy;
Tree[v << 1].lazy += Tree[v].lazy;
Tree[v << 1 | 1].sum += 1LL * Tree[v].lazy * (r - m);
Tree[v << 1 | 1].vmin += Tree[v].lazy;
Tree[v << 1 | 1].vmax += Tree[v].lazy;
Tree[v << 1 | 1].lazy += Tree[v].lazy;
Tree[v].lazy = 0;
}
void modify(int x,int L,int R,int l = 1, int r = n, int v = 1){
if(l >= L && r <= R){
Tree[v].lazy += x;
Tree[v].sum += 1LL * x * (r - l + 1);
Tree[v].vmin += x;
Tree[v].vmax += x;
}
else {
if(Tree[v].lazy)push_down(l,r,v);
int m = (l + r) / 2;
if(L <= m)modify(x,L,R,ls);
if(R > m)modify(x,L,R,rs);
push_up(v);
}
}
node query(int L,int R,int l = 1, int r = n,int v = 1){
if(l >= L && r <= R)
return Tree[v];
else {
if(Tree[v].lazy) push_down(l,r,v);
int m = (l + r) / 2;
node ans;
ans.vmin = oo, ans.vmax = -oo, ans.sum = 0;
if(L <= m){
node nw = query(L,R,ls);
ans.vmin = min(ans.vmin, nw.vmin);
ans.vmax = max(ans.vmax, nw.vmax);
ans.sum += nw.sum;
}
if(R > m){
node nw = query(L,R,rs);
ans.vmin = min(ans.vmin, nw.vmin);
ans.vmax = max(ans.vmax, nw.vmax);
ans.sum += nw.sum;
}
return ans;
}
}
};
SegTree Tr;
int main(){
freopen("redpacket.in","r",stdin);
freopen("redpacket.out","w",stdout);
cin>>n;
for(int i = 1; i < n; i++){
int u,v;
scanf("%d%d",&u,&v);
adde(u,v);
adde(v,u);
}
dep[1] = 1;
dfs(1,1);
// Tr.build();
cin>>m;
for(int i = 1; i <= m; i++){
string opt;
cin>>opt;
if(opt[0] == ‘a‘){
int u;
scanf("%d",&u);
SegTree::node nw;
nw = Tr.query(in[u],out[u]);
cout<<nw.sum<<endl;
}
else if(opt[0] == ‘g‘){
int u, x;
scanf("%d%d",&u,&x);
Tr. modify(x,in[u],out[u]);
}
else {
int u;
scanf("%d",&u);
SegTree::node nw;
nw = Tr.query(in[u],in[u]);
cout<<nw.sum<<endl;
}
}
}
第三题：比如这条红链上＋5，在两个节点上+5，他们的LCA（root）-5，父节点也-5，单点query时就查子树和就好了
#include<bits/stdc++.h>
using namespace std;
const int maxn = 100005, P = 20;
int in[maxn], f[maxn], head[2 * maxn],to[2 * maxn],last[2 *maxn],co[2 * maxn];
int idx, anc[maxn][P+1], dep[maxn], out[maxn], seq[maxn], idx1;
vector <int> G[maxn];
int n,m;
void dfs(int u,int from){
in[u] = ++idx;
anc[u][0] = from;
//fprintf(stderr, "from = %d u = %d\n",from, u);
for(int p = 1; p <= P; p++) {
// fprintf(stderr, "p = %d anc[%d][%d] = %d\n", p, u, p, anc[u][p]);
anc[u][p] = anc[anc[u][p-1]][p-1];
}
for(int i = head[u]; i; i = last[i]){
int v = to[i];
if(v != from){
dep[v] = dep[u] + 1;
f[v] = u;
dfs(v,u);
}
}
out[u] = idx;
}
int lca(int u,int v){//倍增求LCA
if(dep[u] < dep[v])swap(u, v);
int t = dep[u] - dep[v];
for(int p = 0; t; t >>= 1, p++)
if(t & 1)u = anc[u][p];
if(u == v)return v;
for(int p = P; p >=0; p--)
if(anc[u][p] != anc[v][p])
u = anc[u][p], v = anc[v][p];
return anc[u][0];
}
int query(int x){
int ret = 0;
while(x > 0){
ret += seq[x];
x -= (x & -x);
}
return ret;
}
void issum(int x,int val){
while(x <= n){
seq[x] += val;
x += (x & -x);
}
}
void adde(int u,int v){
idx1++;
last[idx1] = head[u];
to[idx1] = v;
head[u] = idx1;
}
int main(){
freopen("redpacket2.in","r",stdin);
freopen("redpacket2.out","w",stdout);
//ios::sync_with_stdio(false);
cin>>n;
for(int i = 1; i < n; i++){
int u,v;
scanf("%d%d",&u,&v);
adde(u,v);
adde(v,u);
}
f[1] = 0;
dep[1] = 1;
dfs(1,1);
cin>>m;
for(int i = 1;i <= m; i++){
string opt;
cin>>opt;
if(opt[0] == ‘s‘){
int u;
scanf("%d",&u);
printf("%d\n",query(out[u])-query(in[u]-1));
}
else{
int u,v,val;
scanf("%d%d%d",&u,&v,&val);
issum(in[u],val);
issum(in[v],val);
int q = lca(u, v);
issum(in[q],-val);
int p = f[q];
if(p != 0)
issum(in[p],-val);
}
}
}
原文地址：https://www.cnblogs.com/EdSheeran/p/8428184.html
时间： 2024-11-09 02:47:15