【jzoj5248】【NOIP2017提高A组模拟8.10】【花花的聚会】【动态规划】【可持久化线段树】

题目大意

解题思路

设f[i]表示i到根最小花费，用可持久化线段树维护到根的路径上的f，区间求最小值即可。

code

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#define LF double#define LL long long#define ULL unsigned int#define fo(i,j,k) for(int i=j;i<=k;i++)#define fd(i,j,k) for(int i=j;i>=k;i--)#define fr(i,j) for(int i=begin[j];i;i=next[i])#define f2(i,j) for(int i=begi2[j];i;i=nex2[i])using namespace std;int const mn=1e5+9,inf=1e9+7;int n,m,q,gra,gr2,pon,begin[mn],begi2[mn],to[mn],next[mn],tim[mn],    cost[mn],nex2[mn],f[mn],mi[mn*20],son[mn*20][2];void insert(int u,int v){    to[++gra]=v;    next[gra]=begin[u];    begin[u]=gra;}void inser2(int v,int k,int w){    tim[++gr2]=k;    cost[gr2]=w;    nex2[gr2]=begi2[v];    begi2[v]=gr2;}int min(int x,int y){    return (x<y)?x:y;}int max(int x,int y){    return (x>y)?x:y;}int qury(int p,int l,int r,int u,int v){    int md=(l+r)/2;    if((l==u)&&(r==v))return mi[p];    if(v<=md)return qury(son[p][0],l,md,u,v);    else if(md<u)return qury(son[p][1],md+1,r,u,v);    else return min(qury(son[p][0],l,md,u,md),qury(son[p][1],md+1,r,md+1,v));}void oper(int p,int q,int l,int r,int u,int v){    int md=(l+r)/2;    if(l==r){        mi[p]=v;        return;    }    if(u<=md){        son[p][1]=son[q][1];        oper(son[p][0]=++pon,son[q][0],l,md,u,v);    }else if(md<u){        son[p][0]=son[q][0];        oper(son[p][1]=++pon,son[q][1],md+1,r,u,v);    }    mi[p]=min(mi[son[p][0]],mi[son[p][1]]);}void dfs(int p,int q,int dep){    if(q)f2(i,p)f[p]=min(f[p],qury(q,1,n,max(dep-tim[i],1),dep-1)+cost[i]);    oper(p,q,1,n,dep,f[p]);    fr(i,p)dfs(to[i],p,dep+1);}int main(){    freopen("d.in","r",stdin);    freopen("d.out","w",stdout);    scanf("%d%d",&n,&m);    fo(i,1,n-1){        int u,v;        scanf("%d%d",&u,&v);        insert(v,u);    }    fo(i,1,m){        int v,k,w;        scanf("%d%d%d",&v,&k,&w);        inser2(v,k,w);    }    pon=n;    mi[0]=inf;    fo(i,2,n)f[i]=inf;    dfs(1,0,1);    scanf("%d",&q);    fo(i,1,q){        int x;        scanf("%d",&x);        printf("%d\n",f[x]);    }    return 0;}