整体二分
先二分一个答案，判断是否可行，把可行的全都扔到左边，不可行的扔到右边
判断是否可行用树状数组就行
具体细节看代码好了
整体二分细节真多……也可能是我大脑已经退化了？
1 //minamoto 2 #include<iostream> 3 #include<cstring> 4 #include<cstdio> 5 #include<vector> 6 #define ll long long 7 using namespace std; 8 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) 9 char buf[1<<21],*p1=buf,*p2=buf; 10 inline ll read(){ 11 #define num ch-'0' 12 char ch;bool flag=0;ll res; 13 while(!isdigit(ch=getc())) 14 (ch=='-')&&(flag=true); 15 for(res=num;isdigit(ch=getc());res=res*10+num); 16 (flag)&&(res=-res); 17 #undef num 18 return res; 19 } 20 const int N=700005; 21 int n,m,k,ans[N],mb[N],st[N],top; 22 ll c[N]; 23 vector<int> a[N]; 24 struct node{ 25 int l,r,op,id;ll k; 26 node(){} 27 node(int l,int r,int op,int id,ll k):l(l),r(r),op(op),id(id),k(k){} 28 }q[N],q1[N],q2[N]; 29 inline void add(int x,ll y){ 30 for(;x<=m;x+=x&-x){ 31 if(!c[x]) st[++top]=x; 32 c[x]+=y; 33 } 34 } 35 ll query(int x){ 36 ll res=0; 37 for(;x;x-=x&-x) res+=c[x]; 38 return res; 39 } 40 void solve(int l,int r,int ql,int qr){ 41 if(l>r) return; 42 if(ql==qr){ 43 for(int i=l;i<=r;++i) 44 if(q[i].op==1) ans[q[i].id]=ql; 45 return; 46 } 47 int mid=ql+qr>>1,p1=0,p2=0; 48 for(int i=l;i<=r;++i){ 49 switch(q[i].op){ 50 case 1:{ 51 ll res=0; 52 for(int j=0,s=a[q[i].id].size();j<s;++j){ 53 res+=query(a[q[i].id][j]); 54 if(res>=q[i].k) break; 55 } 56 if(res>=q[i].k) q1[++p1]=q[i]; 57 else q[i].k-=res,q2[++p2]=q[i]; 58 break; 59 } 60 case 2:{ 61 if(q[i].id<=mid) add(q[i].l,q[i].k),add(q[i].r+1,-q[i].k),q1[++p1]=q[i]; 62 else q2[++p2]=q[i]; 63 break; 64 } 65 case 3:{ 66 if(q[i].id<=mid) add(1,q[i].k),add(q[i].r+1,-q[i].k),add(q[i].l,q[i].k),q1[++p1]=q[i]; 67 else q2[++p2]=q[i]; 68 break; 69 } 70 } 71 } 72 while(top) c[st[top--]]=0; 73 for(int i=1;i<=p1;++i) q[l+i-1]=q1[i]; 74 for(int i=1;i<=p2;++i) q[l+p1+i-1]=q2[i]; 75 solve(l,l+p1-1,ql,mid),solve(l+p1,r,mid+1,qr); 76 } 77 int main(){ 78 // freopen("testdata.in","r",stdin); 79 n=read(),m=read(); 80 for(int i=1;i<=m;++i) a[read()].push_back(i); 81 for(int i=1;i<=n;++i) mb[i]=read(); 82 k=read(); 83 for(int i=1,l,r,t;i<=k;++i){ 84 l=read(),r=read(),t=read(); 85 q[i]=node(l,r,r>=l?2:3,i,t); 86 } 87 for(int i=1;i<=n;++i) 88 q[i+k]=node(0,0,1,i,mb[i]); 89 solve(1,k+n,1,k+1); 90 for(int i=1;i<=n;++i) 91 (ans[i]!=k+1)?printf("%d ",ans[i]):puts("NIE"); 92 return 0; 93 }