HDU 3874 Necklace (树状数组 | 线段树 的离线处理)
Necklace
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2083 Accepted Submission(s): 747
Now Mery thinks the necklace is too long. She plans to take some continuous part of the necklace to build a new one. She wants to know each of the beautiful value of M continuous parts of the necklace. She will give you M intervals [L,R] (1<=L<=R<=N) and you must tell her F(L,R) of them.
For each case, the first line is a number N,1 <=N <=50000, indicating the number of the magic balls. The second line contains N non-negative integer numbers not greater 1000000, representing the beautiful value of the N balls. The third line has a number M, 1 <=M <=200000, meaning the nunber of the queries. Each of the next M lines contains L and R, the query.
6
1 2 3 4 3 5
3
1 2
3 5
2 6
6
1 1 1 2 3 5
3
1 1
2 4
3 5
7
14
1
3
6
- /*
- 题意为查找区间去重后的和
- 用树状数组离线处理
- 将所有查询以右端点从小到大排序
- 按此顺序边去重边查询
- 前面的去重就不会影响到后面的结果了
- */
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm> using namespace std; const int N=;
const int M=; struct node{
int l,r;
int id;
}q[M]; int n,m,val[N],pre[N],loc[];
long long arr[N],res[M]; int lowbit(int x){
return x&(-x);
} void update(int i,int x){
while(i<=n){
arr[i]+=x;
i+=lowbit(i);
}
} long long Sum(int i){
long long ans=;
while(i>){
ans+=arr[i];
i-=lowbit(i);
}
return ans;
} bool cmp(node a,node b){
return a.r<b.r;
} int main(){ //freopen("input.txt","r",stdin); int t;
scanf("%d",&t);
while(t--){
memset(arr,,sizeof(arr));
memset(loc,-,sizeof(loc));
scanf("%d",&n);
for(int i=;i<=n;i++){
scanf("%d",&val[i]);
pre[i]=loc[val[i]];
loc[val[i]]=i;
update(i,val[i]);
}
scanf("%d",&m);
for(int i=;i<=m;i++){
scanf("%d%d",&q[i].l,&q[i].r);
q[i].id=i;
}
sort(q+,q++m,cmp);
int r=;
for(int i=;i<=m;i++){
for(int j=r+;j<=q[i].r;j++)
if(pre[j]!=-)
update(pre[j],-val[j]);
r=q[i].r;
res[q[i].id]=Sum(q[i].r)-Sum(q[i].l-);
}
for(int i=;i<=m;i++)
printf("%I64d\n",res[i]);
}
return ;
}
线段树:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map> using namespace std; const int N=; //#define L(rt) (rt<<1)
//#define R(rt) (rt<<1|1) #define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1 struct Tree{
int l,r;
int id;
}q[N<<]; map<int,int> mp;
int n,m,a[N];
long long sum[N<<],res[N<<]; void PushUp(int rt){
sum[rt]=sum[rt<<]+sum[rt<<|];
} void update(int id,int val,int l,int r,int rt){
if(l==r){
sum[rt]+=val;
return ;
}
int mid=(l+r)>>;
if(id<=mid)
update(id,val,lson);
else
update(id,val,rson);
PushUp(rt);
} long long query(int L,int R,int l,int r,int rt){
if(L<=l && R>=r)
return sum[rt];
int mid=(l+r)>>;
long long ans=;
if(L<=mid)
ans+=query(L,R,lson);
if(R>mid)
ans+=query(L,R,rson);
return ans;
} int cmp(Tree a,Tree b){
return a.r<b.r;
} int main(){ //freopen("input.txt","r",stdin); int t;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i=;i<=n;i++)
scanf("%d",&a[i]);
scanf("%d",&m);
for(int i=;i<=m;i++){
scanf("%d%d",&q[i].l,&q[i].r);
q[i].id=i;
}
sort(q+,q++m,cmp);
mp.clear();
memset(sum,,sizeof(sum));
int r=;
for(int i=;i<=m;i++){
for(int j=r+;j<=q[i].r;j++){
if(mp[a[j]])
update(mp[a[j]],-a[j],,n,);
update(j,a[j],,n,);
mp[a[j]]=j;
r=q[i].r;
}
res[q[i].id]=query(q[i].l,q[i].r,,n,);
}
for(int i=;i<=m;i++)
printf("%I64d\n",res[i]);
}
return ;
}
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