版权声明：欢迎随便转载。 https://blog.csdn.net/a1214034447/article/details/83384033
题目链接:https://darkbzoj.cf/problem/4504
解题思路:
题目跟洛谷P2048 差不多,主要差在区间数只能统计一次,洛谷那题区间一样的数可以重复统计.
那么就可以用pre[i]表示i这个数上一次出现的位置在哪,那么对于新的主席树更新区间就是(pre[i],i),之后的操作与洛谷P2048几乎相同
https://blog.csdn.net/a1214034447/article/details/83246731
#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
#define lson l,mid
#define rson mid+1,r
using namespace std;
typedef long long ll;
const int mx = 2e5 + 10;
const int mod = 1e9+7;
const ll INF = -1e16;
int n,m,root[mx];
int a[mx],rs[mx*30];
int siz,ls[mx*30],d;
ll add[mx*30];
struct node
{
int rt,l,r,p;
ll c;
bool operator < (node A)const
{
return c < A.c;
}
};
struct data
{
int ps;
ll Ma;
}s[mx*30];
map <int,int> mp;
priority_queue <node> st;
void build(int& rt,int l,int r)
{
rt = siz++;
if(l==r){
s[rt].ps = l;
return ;
}
int mid = (l+r)>>1;
build(ls[rt],lson);
build(rs[rt],rson);
}
data up(data a,data b)
{
if(b.Ma>a.Ma) return b;
return a;
}
void update(int x,int &y,int l,int r,int L,int R)
{
y = siz++;
s[y] = s[x],add[y] = add[x];
ls[y] = ls[x],rs[y] = rs[x];
if(L<=l&&r<=R){
add[y] += d;
s[y].Ma += d;
return ;
}
int mid = (l+r)>>1;
if(R<=mid) update(ls[x],ls[y],lson,L,R);
else if(L>mid) update(rs[x],rs[y],rson,L,R);
else{
update(ls[x],ls[y],lson,L,mid);
update(rs[x],rs[y],rson,mid+1,R);
}
s[y] = up(s[ls[y]],s[rs[y]]);
s[y].Ma += add[y];
}
data query(int rt,int l,int r,int L,int R)
{
if(L<=l&&r<=R) return s[rt];
int mid = (l+r)>>1;
data ans = {0,INF};
if(L<=mid) ans = up(ans,query(ls[rt],lson,L,R));
if(R>mid) ans = up(ans,query(rs[rt],rson,L,R));
return data{ans.ps,ans.Ma+add[rt]};
}
int main()
{
scanf("%d%d",&n,&m);
build(root[0],1,n);
for(int i=1;i<=n;i++){
scanf("%d",a+i);
d = a[i];
update(root[i-1],root[i],1,n,mp[a[i]]+1,i);
mp[a[i]] = i;
data ret = query(root[i],1,n,1,i);
st.push(node{root[i],1,i,ret.ps,ret.Ma});
}
data ret;node now;
while(m--){
now = st.top();
st.pop();
if(now.p-now.l>0){
ret = query(now.rt,1,n,now.l,now.p-1);
st.push(node{now.rt,now.l,now.p-1,ret.ps,ret.Ma});
}
if(now.r-now.p>0){
ret = query(now.rt,1,n,now.p+1,now.r);
st.push(node{now.rt,now.p+1,now.r,ret.ps,ret.Ma});
}
//cout << now.c << endl;
}
printf("%lld\n",now.c);
return 0;
}