【JZOJ5295】【清华集训2017模拟】Create

Description

Data Constraint

Solution

这道题很经典。
我们发现一次操作最多使序列多出1段新的连续序列。所以我们考虑维护这个序列。我们将询问按x大小排序后建一棵主席树，那么对于一个修改(l,r,v)，我们可以算出修改后(l,r)对答案的贡献。现在问题是怎样取消修改前的贡献。我们用线段树维护每一段连续的颜色的起点和终点。每一次修改暴力跳一下每个颜色段，查询他们原来的贡献。即可撤销修改前的贡献。时间复杂度O(Qlogm)。

Code

#include<iostream>#include<cmath>#include<cstring>#include<cstdio>#include<algorithm>#include<set>#define ll long longusing 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);    }}