以下模板只是暂时记录一下下，以后会更新学习笔记
树剖
void dfs1(int n,int F)
{
s[n]=1;//初始化子树大小为1
d[n]=d[F]+1;
f[n]=F;
int V;
for(int i=head[n];i;i=e[i].next)
{
V=e[i].v;
if(V!=F)
{
dfs1(V,n);
s[n]+=s[V];
if(s[son[n]]<s[V]) son[n]=V;
}
}
}
void dfs2(int n,int C)
{
c[n]=C;
if(son[n]) dfs2(son[n],C);
else return;
int V;
for(int i=head[n];i;i=e[i].next)
{
V=e[i].v;
if(V!=f[n] && V!=son[n]) dfs2(V,V);
}
}
树链剖分求lca
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<string>
#include<cstring>
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
const int N=500001;
int n,m,root,cnt;
struct node
{
int u,v,next;
}e[N<<1];
int head[N];
int son[N];
int f[N],d[N],s[N],c[N];
void add(int x,int y)
{
e[++cnt].u=x;
e[cnt].v=y;
e[cnt].next=head[x];
head[x]=cnt;
}
void dfs1(int n,int F)
{
s[n]=1;//初始化子树大小为1
d[n]=d[F]+1;
f[n]=F;
int V;
for(int i=head[n];i;i=e[i].next)
{
V=e[i].v;
if(V!=F)
{
dfs1(V,n);
s[n]+=s[V];
if(s[son[n]]<s[V]) son[n]=V;
}
}
}
void dfs2(int n,int C)
{
c[n]=C;
if(son[n]) dfs2(son[n],C);
else return;
int V;
for(int i=head[n];i;i=e[i].next)
{
V=e[i].v;
if(V!=f[n] && V!=son[n]) dfs2(V,V);
}
}
int lca(int x,int y)
{
for(;c[x]!=c[y];x=f[c[x]])
{
if(d[c[x]]<d[c[y]]) swap(x,y);
}
return d[x]<d[y]? x: y;
}
int main()
{
cin>>n>>m>>root;
for(int i=1;i<n;++i)
{
int x,y;
cin>>x>>y;
add(x,y);
add(y,x);
}
dfs1(root,0);
dfs2(root,root);
for(int i=1;i<=m;++i)
{
int x,y;
cin>>x>>y;
cout<<lca(x,y)<<'
';
}
}
无向图如何判断一个点c是否在某两点u,v间的最短路上?
(dis_{u,c}+dis_{c,v}=dis_{u,v})
三分法模板
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<string>
#include<cstring>
using namespace std;
#define eps 0.000000001
inline int read() {
char c = getchar();
int x = 0, f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
const int N=10l;
int n;
double l,r;
double a[N];
double calc(double x)
{
double ans=0;
for(int i=n;i>=0;--i)
{
ans+=a[i]*pow(x,i);
}
return ans;
}
int main()
{
scanf("%d%lf%lf",&n,&l,&r);
for(int i=n;i>=0;i--)scanf("%lf",&a[i]);
while(r-l>eps){
double mid1=(2*l+r)/3.0,mid2=(l+2*r)/3.0;
if(calc(mid1)<calc(mid2))l=mid1;
else r=mid2;
}
printf("%.5lf",l);
return 0;
}
倍增求LCA
#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 500010
using namespace std;
int n,q,root,fa[maxn][22],dep[maxn],num,head[maxn];
struct node{int to,pre;}e[maxn*2];
void Insert(int from,int to){
e[++num].to=to;
e[num].pre=head[from];
head[from]=num;
}
void dfs(int now,int father){
fa[now][0]=father;
dep[now]=dep[father]+1;
for(int i=head[now];i;i=e[i].pre){
int to=e[i].to;
if(to==father)continue;
dfs(to,now);
}
}
int get(int a,int delta){
for(int i=0;i<=21;i++){
if(delta&(1<<i))a=fa[a][i];
}return a;
}
int lca(int a,int b){
if(dep[a]<dep[b])swap(a,b);
a=get(a,dep[a]-dep[b]);
if(a==b) return a;
for(int i=21;i>=0;i--)
if(fa[a][i]!=fa[b][i])
a=fa[a][i],b=fa[b][i];
return fa[a][0];
}
int main(){
scanf("%d%d%d",&n,&q,&root);
int x,y;
for(int i=1;i<n;i++){
scanf("%d%d",&x,&y);
Insert(x,y);Insert(y,x);
}
dfs(root,root);
for(int j=1;j<=20;j++)
for(int i=1;i<=n;i++)
fa[i][j]=fa[fa[i][j-1]][j-1];
while(q--){
scanf("%d%d",&x,&y);
printf("%d
",lca(x,y));
}
}
线段树5种操作模板
#include<cstdio>
using namespace std;
int n,p,a,b,m,x,y,ans;
struct node
{
int l,r,value,lazy;
}tree[400001];
inline void build(int k,int ll,int rr)//建树
{
tree[k].l=ll,tree[k].r=rr;
if(tree[k].l==tree[k].r)
{
scanf("%d",&tree[k].value);
return;
}
int m=(ll+rr)/2;
build(k*2,ll,m);
build(k*2+1,m+1,rr);
tree[k].value=tree[k*2].value+tree[k*2+1].value;
}
inline void pushdown(int k)//标记下传
{
tree[k*2].lazy+=tree[k].lazy;
tree[k*2+1].lazy+=tree[k].lazy;
tree[k*2].value+=tree[k].lazy*(tree[k*2].r-tree[k*2].l+1);
tree[k*2+1].value+=tree[k].lazy*(tree[k*2+1].r-tree[k*2+1].l+1);
tree[k].lazy=0;
}
inline void ask_point(int k)//单点查询
{
if(tree[k].l==tree[k].r)
{
ans=tree[k].value;
return ;
}
if(tree[k].lazy) pushdown(k);
int m=(tree[k].l+tree[k].r)/2;
if(x<=m) ask_point(k*2);
else ask_point(k*2+1);
}
inline void change_point(int k)//单点修改
{
if(tree[k].l==tree[k].r)
{
tree[k].value+=y;
return;
}
if(tree[k].lazy) pushdown(k);
int m=(tree[k].l+tree[k].r)/2;
if(x<=m) change_point(k*2);
else change_point(k*2+1);
tree[k].value=tree[k*2].value+tree[k*2+1].value;
}
inline void ask_interval(int k)//区间查询
{
if(tree[k].l>=a&&tree[k].r<=b)
{
ans+=tree[k].value;
return;
}
if(tree[k].lazy) pushdown(k);
int m=(tree[k].l+tree[k].r)/2;
if(a<=m) ask_interval(k*2);
if(b>m) ask_interval(k*2+1);
}
inline void update(int k)//区间修改
{
if(tree[k].l>=a&&tree[k].r<=b)
{
tree[k].value+=(tree[k].r-tree[k].l+1)*y;
tree[k].lazy+=y;
return;
}
if(tree[k].lazy) pushdown(k);
int m=(tree[k].l+tree[k].r)/2;
if(a<=m) update(k*2);
if(b>m) update(k*2+1);
tree[k].value=tree[k*2].value+tree[k*2+1].value;
}
int main()
{
scanf("%d",&n);//n个节点
build(1,1,n);//建树
scanf("%d",&m);//m种操作
for(int i=1;i<=m;i++)
{
scanf("%d",&p);
ans=0;
if(p==1)
{
scanf("%d",&x);
ask_point(1);//单点查询,输出第x个数
printf("%d",ans);
}
else if(p==2)
{
scanf("%d%d",&x,&y);
change_point(1);//单点修改
}
else if(p==3)
{
scanf("%d%d",&a,&b);//区间查询
ask_interval(1);
printf("%d
",ans);
}
else
{
scanf("%d%d%d",&a,&b,&y);//区间修改
update(1);
}
}
}