斜率优化
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struct CHT{
int q[maxn]; //存决策点标号
ll X(int pos){/* */}
ll Y(int pos){/* */}
double get_K(int x,int y){
return (double)(Y(x)-Y(y))/(double)(X(x)-X(y));
}
int get_pos(int l,int r,int k){ //二分找到决策点
int ans=l;l++;
while(r>=l){
int mid=(l+r)>>1;
if(get_K(q[mid],q[mid-1])<=k)ans=mid,l=mid+1; //维护下凸包
else r=mid-1;
}
return q[ans];
}
}T;
1. [HNOI2008]玩具装箱TOY
斜率优化
设\(dp_i\)表示前\(i\)件玩具的最小费用，\(sum_i=\sum_{1\leq j\leq i}c_j\)
\(dp_i=min\{dp_j+[i-(j+1)+sum_i-sum_j-L]^2 \}\)
设\(a_i=i+sum_i，b_j=j+1+sum_j+L\)
\(dp_i=min\{ dp_j+a_i^2-2*a_i*b_j+b_j^2 \}\)
\(dp_j+b_j^2=2*a_i*b_j+dp_i-a_i^2\)
形如\(Y=k*X+b\)
\(Y=dp_j+b_j^2，X=b_j，k=2*a_i\)
一般\(X\)和\(Y\)都是决策状态，也就是与\(j\)相关的，我们要求的\(dp_i\)放在\(b\)里，这样我们只要求截距最小的就是当前\(dp_i\)的最小值
也就从之前的所有\(（X,Y）\)中选一个点，使得经过那个点的直线截距最小，那么我们维护下凸包，然后二分找到那个点
因为这道题的每个i的斜率是单增的，也就是\(k=2*a_i\)是单增的，那么这道题我们也可以不用二分，直接维护队首的元素，把不优的直接弹出
点击查看代码
#include <bits/stdc++.h>
#define ll long long
#define pa pair<int,int>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define YES {puts("YES");return;}
#define NO {puts("NO");return ;}
using namespace std;
const int maxn=2e5+101;
const int MOD=1e9+7;
const ll inf=2147483647;
const double eps=1e-12;
ll read(){
ll x=0,f=1;char ch=getchar();
for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
return x*f;
}
ll n,L,c[maxn],sum[maxn],dp[maxn],a[maxn],b[maxn];
struct CHT{
int q[maxn]; //存决策点标号
ll X(int pos){return b[pos];}
ll Y(int pos){return dp[pos]+b[pos]*b[pos];}
double get_K(int x,int y){
return (double)(Y(x)-Y(y))/(double)(X(x)-X(y));
}
int get_pos(int l,int r,int k){ //二分找到决策点
int ans=l;l++;
while(r>=l){
int mid=(l+r)>>1;
if(get_K(q[mid],q[mid-1])<=k)ans=mid,l=mid+1;
else r=mid-1;
}
return q[ans];
}
}T;
int main(){
n=read();L=read();b[0]=L+1;
for(int i=1;i<=n;i++){
c[i]=read();
sum[i]=sum[i-1]+c[i];
a[i]=sum[i]+i;
b[i]=i+1+sum[i]+L;
}
int tail=1,head=1;dp[1]=(a[1]-b[0])*(a[1]-b[0]);
for(int i=2;i<=n;i++){
int id=T.get_pos(head,tail,2*a[i]);
dp[i]=dp[id]+(a[i]-b[id])*(a[i]-b[id]);
while(head<tail && T.get_K(T.q[tail],T.q[tail-1])>T.get_K(T.q[tail-1],i))tail--;
T.q[++tail]=i;
}
printf("%lld\n",dp[n]);
return 0;
}
点击查看代码
#include <bits/stdc++.h>
#define ll long long
#define pa pair<ll,ll>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define YES {puts("YES");return;}
#define NO {puts("NO");return ;}
using namespace std;
const int maxn=2e5+101;
const int MOD=1e9+7;
const ll inf=1e17;
const double eps=1e-12;
ll read(){
ll x=0,f=1;char ch=getchar();
for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
return x*f;
}
ll n,m,K;
ll dp[21][maxn];
struct CHT{
ll q[maxn]; //存决策点标号
ll now; //存第几层
ll X(ll pos){return pos;}
ll Y(ll pos){return dp[now][pos]+pos*pos;}
double get_K(ll x,ll y){
return (double)(Y(x)-Y(y))/(double)(X(x)-X(y));
}
int get_pos(ll l,ll r,ll k){ //二分找到决策点
ll ans=l;l++;
while(r>=l){
ll mid=(l+r)>>1;
if(get_K(q[mid],q[mid-1])<=k)ans=mid,l=mid+1;
else r=mid-1;
}
return q[ans];
}
}T;
vector<vector<pa> >G(maxn+1);
int vis[maxn];
struct node{
ll dis,pos;
bool operator <(const node &x)const{
return x.dis<dis;
}
};
void Dij(int now){
for(int i=1;i<=n;i++)vis[i]=0;
priority_queue<node>que;
for(int i=1;i<=n;i++){
que.push({dp[now][i],i});
}
while(que.size()){
auto x=que.top();que.pop();
ll u=x.pos,dis=x.dis;
if(vis[u])continue;
vis[u]=1;
for(auto ed:G[u]){
ll v=ed.fi,w=ed.se;
if(dp[now][v]>dis+w){
dp[now][v]=dis+w;
if(!vis[v])que.push({dp[now][v],v});
}
}
}
return ;
}
int main(){
n=read();m=read();K=read();
for(int i=1;i<=m;i++){
int x=read(),y=read(),z=read();
G[x].pb(mp(y,z));
G[y].pb(mp(x,z));
}
for(int k=0;k<=K;k++)for(int u=1;u<=n;u++)dp[k][u]=1e16;
dp[0][1]=0;Dij(0);
int tail,head;
for(int k=1;k<=K;k++){
T.now=k-1;head=1;tail=0;
for(ll u=1;u<=n;u++){
while(tail>head && T.get_K(T.q[tail-1],T.q[tail])>=T.get_K(u,T.q[tail-1]))tail--;
T.q[++tail]=u;
}
for(ll u=1;u<=n;u++){
ll v=T.get_pos(head,tail,2*u);
dp[k][u]=min(dp[k][u],dp[k-1][v]+(u-v)*(u-v));
}
Dij(k);
}
for(int i=1;i<=n;i++){
ll ans=inf;
for(int j=0;j<=K;j++)ans=min(ans,dp[j][i]);
cout<<ans<<" ";
}
return 0;
}
