2019 Nanchang Online Contest Problem I Yukino With Subinterval

将每个连续的最右端赋值权值1，需要写带修改的主席树，我没写过拿CDQ分治水过

这题居然是全场AC最少的题目，我一个半小时搞定（其实是因为写主席树写了一半发现不会写，改成了CDQ分治）

#include<bits/stdc++.h>
using namespace std;
#define maxn 2000005
int s,w,n,cnt;
int ans[maxn],sum[maxn],A[maxn];
struct data{int x,y,z,pos,id;}a[maxn],b[maxn];
inline bool cmp(data a,data b){if (a.x==b.x&&a.y==b.y) return a.pos<b.pos;else if (a.x==b.x) return a.y<b.y;else return a.x<b.x;
}
inline void add(int x,int v){for(int i=x;i<=w;i+=i&(-i)) sum[i]+=v;
}
inline int query(int x){int ret=0;for(int i=x;i;i-=i&(-i)) ret+=sum[i];return ret;
}
inline void solve(int l,int r){if (l==r) return;int mid=(l+r)/2,l1=l,l2=mid+1; for(int i=l;i<=r;++i){if (a[i].id<=mid&&!a[i].pos) add(a[i].y,a[i].z);if (a[i].id>mid&&a[i].pos) ans[a[i].pos]+=query(a[i].y)*a[i].z;}for(int i=l;i<=r;++i) if (a[i].id<=mid&&!a[i].pos) add(a[i].y,-a[i].z);for(int i=l;i<=r;++i){if (a[i].id<=mid) b[l1++]=a[i];else b[l2++]=a[i];}for(int i=l;i<=r;++i) a[i]=b[i];solve(l,mid);solve(mid+1,r);
}
void add(int x,int y,int z){n++;a[n]=(data){x,y,z,0,n};}
int main(){s=0;int m,x,y,z;cin>>w>>m;for(int i=1;i<=w;++i)scanf("%d",&A[i]);for(int i=1;i<=w;++i){if(A[i]==A[i-1]){x=i-1;y=A[i-1];z=-1;n++;a[n]=(data){x,y,z,0,n};}x=i;y=A[i];z=1;n++;a[n]=(data){x,y,z,0,n};}while(m--){int op;scanf("%d",&op);if (op==1){scanf("%d%d",&x,&y);if(A[x]==y)continue;if(A[x]==A[x-1]&&A[x]==A[x+1]){add(x-1,A[x-1],1);add(x,y,1);}else if(A[x]==A[x-1]&&A[x]!=A[x+1]){add(x-1,A[x-1],1);add(x,A[x],-1);if(A[x+1]!=y)add(x,y,1);}else if(A[x]!=A[x-1]&&A[x]==A[x+1]){if(A[x-1]==y){add(x-1,A[x-1],-1);}add(x,A[x],-1);add(x,y,1);}else if(A[x]!=A[x-1]&&A[x]!=A[x+1]){if(y==A[x-1]&&y==A[x+1]){add(x-1,A[x-1],-1);add(x,A[x],-1);}if(y==A[x-1]&&y!=A[x+1]){add(x-1,A[x-1],-1);add(x,A[x],-1);add(x,y,1);}if(y==A[x+1]&&y!=A[x-1]){add(x,A[x],-1);}if(y!=A[x-1]&&y!=A[x+1]){add(x,A[x],-1);add(x,y,1);}}A[x]=y;}else {int x1,x2,y1,y2;scanf("%d%d%d%d",&x1,&x2,&y1,&y2);--x1,--y1;ans[++cnt]=(x2-x1)*(y2-y1)*s;n++;a[n]=(data){x1,y1,1,cnt,n};n++;a[n]=(data){x1,y2,-1,cnt,n};n++;a[n]=(data){x2,y1,-1,cnt,n};n++;a[n]=(data){x2,y2,1,cnt,n};}}sort(a+1,a+n+1,cmp);solve(1,n);for(int i=1;i<=cnt;i++) printf("%d\n",ans[i]);
}

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