BZOJ3211: 花神游历各国

题解:直接维护最小值　然后裸线段树即可

/**************************************************************
    Problem: 3211
    User: c20161007
    Language: C++
    Result: Accepted
    Time:1492 ms
    Memory:8332 kb
****************************************************************/
 
#include <bits/stdc++.h>
#define ll long long
const int MAXN=1e5+10;
using namespace std;
ll sum[MAXN<<2],key[MAXN],minn[MAXN<<2];
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return f*x;
}
void up(int x){
    sum[x]=sum[x<<1]+sum[x<<1|1];
    minn[x]=max(minn[x<<1],minn[x<<1|1]);
}
void built(int rt,int l,int r){
    if(l==r){sum[rt]=minn[rt]=key[l];return ;}
    int mid=(l+r)>>1;
    built(rt<<1,l,mid);
    built(rt<<1|1,mid+1,r);
    up(rt);
}
void update(int rt,int l,int r,int ql,int qr){
   // cout<<minn[rt]<<endl;
    if(minn[rt]<=1)return ;
    //cout<<l<<" "<<r<<" "<<minn[rt]<<endl;
    if(l==r){key[l]=(ll)sqrt(key[l]);sum[rt]=minn[rt]=key[l];return ;}
    int mid=(l+r)>>1;
    if(ql<=mid)update(rt<<1,l,mid,ql,qr);
    if(qr>mid)update(rt<<1|1,mid+1,r,ql,qr);
    up(rt);
}
ll ans;
void querty(int rt,int l,int r,int ql,int qr){
    if(ql<=l&&r<=qr){ans+=sum[rt];return ;}
    int mid=(l+r)>>1;
    if(ql<=mid)querty(rt<<1,l,mid,ql,qr);
    if(qr>mid)querty(rt<<1|1,mid+1,r,ql,qr);
}
int n,q;
int main(){
    n=read();
    for(int i=1;i<=n;i++)key[i]=read();
    built(1,1,n);
    q=read();int op,l,r;
    for(int i=1;i<=q;i++){
    op=read();l=read();r=read();
    //cout<<l<<" "<<r<<endl;
    if(op==1){ans=0;querty(1,1,n,l,r);printf("%lld\n",ans);}
    else update(1,1,n,l,r);
    }
}

 

3211: 花神游历各国

Time Limit: 5 Sec  Memory Limit: 128 MB
Submit: 5216  Solved: 1915
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Description

 

 

Input

 

 

Output

每次x=1时，每行一个整数，表示这次旅行的开心度

Sample Input

4

1 100 5 5

5

1 1 2

2 1 2

1 1 2

2 2 3

1 1 4

Sample Output

101

11

11

HINT

对于100%的数据， n ≤ 100000，m≤200000 ,data[i]非负且小于10^9