【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 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);
}
}