L - Ch’s gift HDU - 6162
Ch’s gift
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2534 Accepted Submission(s): 887
题目链接
http://acm.hdu.edu.cn/showproblem.php?pid=6162
Problem Description
Cui is working off-campus and he misses his girl friend very much.
After a whole night tossing and turning, he decides to get to his girl
friend's city and of course, with well-chosen gifts. He knows neither
too low the price could a gift be since his girl friend won't like it,
nor too high of it since he might consider not worth to do. So he will
only buy gifts whose price is between [a,b].
There are n cities in
the country and (n-1) bi-directional roads. Each city can be reached
from any other city. In the ith city, there is a specialty of price ci
Cui could buy as a gift. Cui buy at most 1 gift in a city. Cui starts
his trip from city s and his girl friend is in city t. As mentioned
above, Cui is so hurry that he will choose the quickest way to his girl
friend(in other words, he won't pass a city twice) and of course, buy as
many as gifts as possible. Now he wants to know, how much money does he
need to prepare for all the gifts?
Input
For each case:
The first line contains tow integers n,m(1≤n,m≤10^5), representing the number of cities and the number of situations.
The second line contains n integers c1,c2,...,cn(1≤ci≤10^9), indicating the price of city i's specialty.
Then n-1 lines follows. Each line has two integers x,y(1≤x,y≤n), meaning there is road between city x and city y.
Next
m line follows. In each line there are four integers
s,t,a,b(1≤s,t≤n;1≤a≤b≤10^9), which indicates start city, end city, lower
bound of the price, upper bound of the price, respectively, as the
exact meaning mentioned in the description above
Output
Sample Input
Sample Output
Source
题意
题解
AC代码
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 100050
#define INF 123456789
int n,m;
int tot,last[N];
ll ans[N];
int cnt,fa[N],dp[N],size[N],son[N],rk[N],kth[N],top[N];
struct Query
{
int l,r,id; ll val;
bool operator <(const Query&b)const
{return val<b.val;}
}a[N],que[N<<];
struct Edge{int from,to,s;}edges[N<<];
struct Tree{int l,r;ll sum;}tr[N<<];
template<typename T>void read(T&x)
{
ll k=; char c=getchar();
x=;
while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
if (c==EOF)exit();
while(isdigit(c))x=x*+c-'',c=getchar();
x=k?-x:x;
}
void read_char(char &c)
{while(!isalpha(c=getchar())&&c!=EOF);}
void AddEdge(int x,int y)
{
edges[++tot]=Edge{x,y,last[x]};
last[x]=tot;
}
void dfs1(int x,int pre)
{
fa[x]=pre;
dp[x]=dp[pre]+;
size[x]=;
son[x]=;
for(int i=last[x];i;i=edges[i].s)
{
Edge &e=edges[i];
if (e.to==pre)continue;
dfs1(e.to,x);
size[x]+=size[e.to];
if (size[e.to]>size[son[x]])son[x]=e.to;
}
}
void dfs2(int x,int y)
{
rk[x]=++cnt;
kth[cnt]=x;
top[x]=y;
if (son[x]==)return;
dfs2(son[x],y);
for(int i=last[x];i;i=edges[i].s)
{
Edge &e=edges[i];
if (e.to==fa[x]||e.to==son[x])continue;
dfs2(e.to,e.to);
}
}
void bt(int x,int l,int r)
{
tr[x].l=l; tr[x].r=r; tr[x].sum=;
if (l==r)return;
int mid=(l+r)>>;
bt(x<<,l,mid);
bt(x<<|,mid+,r);
}
void update(int x,int p,ll tt)
{
if (p<=tr[x].l&&tr[x].r<=p)
{
tr[x].sum+=tt;
return;
}
int mid=(tr[x].l+tr[x].r)>>;
if (p<=mid)update(x<<,p,tt);
if (mid<p)update(x<<|,p,tt);
tr[x].sum=tr[x<<].sum+tr[x<<|].sum;
}
ll query(int x,int l,int r)
{
if (l<=tr[x].l&&tr[x].r<=r)
return tr[x].sum;
int mid=(tr[x].l+tr[x].r)>>; ll ans=;
if (l<=mid)ans+=query(x<<,l,r);
if (mid<r)ans+=query(x<<|,l,r);
return ans;
}
ll get_sum(int x,int y)
{
int fx=top[x],fy=top[y];ll ans=;
while(fx!=fy)
{
if (dp[fx]<dp[fy])swap(x,y),swap(fx,fy);
ans+=query(,rk[fx],rk[x]);
x=fa[fx]; fx=top[x];
}
if (dp[x]<dp[y])swap(x,y);
ans+=query(,rk[y],rk[x]);
return ans;
}
void work()
{
read(n); read(m);
for(int i=;i<=n;i++)read(a[i].val),a[i].id=i;
for(int i=;i<=n-;i++)
{
int x,y;
read(x); read(y);
AddEdge(x,y);
AddEdge(y,x);
}
int num=;
for(int i=;i<=m;i++)
{
int l,r,x,y;
read(l); read(r); read(x);read(y);
que[++num]=Query{l,r,-i,x-};
que[++num]=Query{l,r,i,y};
}
sort(a+,a+n+);
sort(que+,que+num+);
dfs1(,);
dfs2(,);
bt(,,n);
int ds=;
for(int i=;i<=num;i++)
{
while(ds<=n&&a[ds].val<=que[i].val)
{
update(,rk[a[ds].id],a[ds].val);
ds++;
}
ll sum=get_sum(que[i].l,que[i].r);
if (que[i].id<) ans[-que[i].id]-=sum;
else ans[que[i].id]+=sum;
}
printf("%lld",ans[]);
for(int i=;i<=m;i++)printf(" %lld",ans[i]);
printf("\n");
}
void clear()
{
tot=; cnt=;
memset(last,,sizeof(last));
memset(ans,,sizeof(ans));
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("aa.in","r",stdin);
//freopen("my.out","w",stdout);
#endif
while()
{
clear();
work();
}
}
TLE代码(树链剖分+主席树)
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 100050
#define INF 123456789
int n,m,w[N];ll b[N];
int tot,last[N];
int tree_num,root[N];
int cnt,fa[N],dp[N],size[N],son[N],rk[N],kth[N],top[N];
struct Edge{int from,to,s;}edges[N<<];
struct Tree{int l,r,ls,rs;ll sum;}tr[];
template<typename T>void read(T&x)
{
ll k=; char c=getchar();
x=;
while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
if (c==EOF)exit();
while(isdigit(c))x=x*+c-'',c=getchar();
x=k?-x:x;
}
void read_char(char &c)
{while(!isalpha(c=getchar())&&c!=EOF);}
void AddEdge(int x,int y)
{
edges[++tot]=Edge{x,y,last[x]};
last[x]=tot;
}
void dfs1(int x,int pre)
{
fa[x]=pre;
dp[x]=dp[pre]+;
size[x]=;
son[x]=;
for(int i=last[x];i;i=edges[i].s)
{
Edge &e=edges[i];
if (e.to==pre)continue;
dfs1(e.to,x);
size[x]+=size[e.to];
if (size[e.to]>size[son[x]])son[x]=e.to;
}
}
void dfs2(int x,int y)
{
rk[x]=++cnt;
kth[cnt]=x;
top[x]=y;
if (son[x]==)return;
dfs2(son[x],y);
for(int i=last[x];i;i=edges[i].s)
{
Edge &e=edges[i];
if (e.to==fa[x]||e.to==son[x])continue;
dfs2(e.to,e.to);
}
}
void bt(int &x,int l,int r)
{
x=++tree_num;
tr[x].l=l; tr[x].r=r; tr[x].sum=;
if (l==r)return;
int mid=(l+r)>>;
bt(tr[x].ls,l,mid);
bt(tr[x].rs,mid+,r);
}
void add(int &x,int last,int p)
{
x=++tree_num;
tr[x]=tr[last];
tr[x].sum+=b[p];
if (tr[x].l==tr[x].r)return;
int mid=(tr[x].l+tr[x].r)>>;
if(p<=mid)add(tr[x].ls,tr[last].ls,p);
else add(tr[x].rs,tr[last].rs,p);
}
ll ask(int x,int y,int p)
{
if (tr[x].r<=p)return tr[y].sum-tr[x].sum;
int mid=(tr[x].l+tr[x].r)>>;ll ans=;
if (<=mid)ans+=ask(tr[x].ls,tr[y].ls,p);
if (mid<p)ans+=ask(tr[x].rs,tr[y].rs,p);
return ans;
}
ll get_sum(int x,int y,int tt)
{
int fx=top[x],fy=top[y];ll ans=;
while(fx!=fy)
{
if (dp[fx]<dp[fy])swap(x,y),swap(fx,fy);
ans+=ask(root[rk[fx]-],root[rk[x]],tt);
x=fa[fx]; fx=top[x];
}
if (dp[x]<dp[y])swap(x,y);
ans+=ask(root[rk[y]-],root[rk[x]],tt);
return ans;
}
void work()
{
read(n); read(m);
int num=;
for(int i=;i<=n;i++)read(w[i]),b[++num]=w[i];
b[++num]=INF;
for(int i=;i<=n-;i++)
{
int x,y;
read(x); read(y);
AddEdge(x,y);
AddEdge(y,x);
}
sort(b+,b+num+);
num=unique(b+,b+num+)-b-;
dfs1(,);
dfs2(,);
bt(root[],,num);
for(int i=;i<=n;i++)
{
int tt=lower_bound(b+,b+num+,w[kth[i]])-b;
add(root[i],root[i-],tt);
}
for(int i=;i<=m;i++)
{
if (i>)printf(" ");
int x,y,l,r;
read(x); read(y); read(l); read(r);
l=lower_bound(b+,b+num+,l)-b-;
r=upper_bound(b+,b+num+,r)-b-;
ll ans=get_sum(x,y,r);
ans-=get_sum(x,y,l);
printf("%lld",ans);
}
printf("\n");
}
void clear()
{
tot=; cnt=; tree_num=;
memset(last,,sizeof(last));
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("aa.in","r",stdin);
#endif
while()
{
clear();
work();
}
}
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