【JZOJ5295】【清華集訓2017模擬】Create

Description

Data Constraint

Solution

這道題很經典。
我們發現一次操作最多使序列多出1段新的連續序列。所以我們考慮維護這個序列。我們將詢問按x大小排序後建一棵主席樹，那麼對於一個修改(l,r,v)，我們可以算出修改後(l,r)對答案的貢獻。現在問題是怎樣取消修改前的貢獻。我們用線段樹維護每一段連續的顏色的起點和終點。每一次修改暴力跳一下每個顏色段，查詢他們原來的貢獻。即可撤銷修改前的貢獻。時間複雜度O(Qlogm)。

Code

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<set>
#define ll long long
using namespace std;
const ll maxn=1e5+5;
struct code{
    ll l,r,num,bz;
    bool friend operator < (code x,code y){
        return x.l<y.l;
    }
}f[maxn*60];
struct code1{
    ll l,r,x;
}b[maxn];
ll r[maxn],l[maxn],a[maxn],d[maxn],g[maxn*4],g1[maxn*4],ans1[maxn];
ll n,m,q,i,t,j,k,x,y,z,num,ans,xx,yy,p,qq;
bool cmp(code1 x,code1 y){
    return x.x<y.x;
}
void insert(ll l,ll r,ll &v,ll x,ll y){
    ll mid=(l+r)/2;
    f[++num]=f[v];v=num;
    if (l>=x && r<=y){
        f[v].bz++;f[v].num+=r-l+1;return;
    }
    if (mid>=x) insert(l,mid,f[v].l,x,y);
    if (mid<y) insert(mid+1,r,f[v].r,x,y);
    f[v].num=(r-l+1)*f[v].bz+f[f[v].l].num+f[f[v].r].num;
}
void find(ll l,ll r,ll v,ll x,ll y){
    ll mid=(l+r)/2;
    if (!f[v].num)return;
    if (l>=x && r<=y){
        k+=f[v].num;
        return;
    }else k+=(min(r,y)-max(x,l)+1)*f[v].bz;
    if (mid>=x) find(l,mid,f[v].l,x,y);
    if (mid<y) find(mid+1,r,f[v].r,x,y);
}
void build(ll l2,ll rr,ll v){
    ll mid=(l2+rr)/2;
    if (l2==rr){
        g[v]=l[l2];g1[v]=r[l2];
        return;
    }
    build(l2,mid,v*2);build(mid+1,rr,v*2+1);
}
void find1(ll l,ll r,ll v,ll x){
    ll mid=(l+r)/2;
    if (l==r){
        p=g[v];q=g1[v];return;
    }
    if (g[v]) g[v*2]=g[v*2+1]=g[v],g[v]=0;
    if (g1[v]) g1[v*2]=g1[v*2+1]=g1[v],g1[v]=0;
    if (x>mid) find1(mid+1,r,v*2+1,x);
    else find1(l,mid,v*2,x);
}
void change(ll l,ll r,ll v,ll x,ll y){
    if (x>y) return;
    ll mid=(l+r)/2;
    if (l>=x && r<=y){
        g[v]=x;g1[v]=y;return;
    }
    if (g[v]) g[v*2]=g[v*2+1]=g[v],g[v]=0;
    if (g1[v]) g1[v*2]=g1[v*2+1]=g1[v],g1[v]=0;
    if (mid<y) change(mid+1,r,v*2+1,x,y);
    if (mid>=x) change(l,mid,v*2,x,y);
}
ll pan(ll x){
    ll l=0,r=m,mid;
    while (l<r){
        mid=(l+r+1)/2;
        if (b[mid].x<=x) l=mid;
        else r=mid-1;
    }
    return l;
}
int main(){
    freopen("create.in","r",stdin);freopen("create.out","w",stdout);
    scanf("%lld%lld%lld",&n,&m,&qq);
    for (i=1;i<=n;i++){
        scanf("%lld",&a[i]);
        if (a[i]==a[i-1]) l[i]=l[i-1];
        else l[i]=i;
    }
    for (i=n;i>=1;i--)
        if (a[i]==a[i+1]) r[i]=r[i+1];
        else r[i]=i;
    build(1,n,1);
    for (i=1;i<=m;i++)
        scanf("%lld%lld%lld",&b[i].l,&b[i].r,&b[i].x);
    sort(b+1,b+m+1,cmp);
    for (i=1;i<=m;i++){
        d[i]=d[i-1];
        insert(1,n,d[i],b[i].l,b[i].r);
    }
    i=1;
    while (i<=n){
        t=pan(a[i]);k=0;
        find(1,n,d[t],i,r[i]);
        i=r[i]+1,ans+=k;
    }
    printf("%lld\n",ans);
    for (i=1;i<=qq;i++){
        scanf("%lld%lld%lld",&x,&y,&z);x^=ans;y^=ans;z^=ans;
        k=0;t=pan(z);
        find(1,n,d[t],x,y);ans+=k;
        xx=x;find1(1,n,1,xx);
        change(1,n,1,p,xx-1);
        while (xx<=y){
            find1(1,n,1,xx);
            k=0;t=pan(a[p]);
            find(1,n,d[t],xx,min(q,y));ans-=k;
            xx=min(q,y)+1;
        }
        change(1,n,1,y+1,q);if (y+1<=q)a[y+1]=a[p];
        change(1,n,1,x,y);a[x]=z;
        printf("%lld\n",ans);
    }
}