Luogu P3527 [POI2011]MET-Meteors 整体二分

思路：整体二分

提交：4次

错因：树状数组开的$int$

题解：

二分操作序列，将仅用$[l,md]$即可满足要求的国家递归到左半边，将仅用$[l,md]$不能满足要求的国家，把他们的要求去掉左半边的贡献，递归到右半边。

具体来说，开一个以空间站为下标的树状数组（把环展成链），区间加单点求和转化为差分和前缀和，依次加入$[l,md]$中的所有操作区间；

然后每个国家枚举自己的所有空间站，计算贡献，判断前$[l,md]$是否满足，来决定向左右递归的方向。

#include<iostream>
#include<cstdio>
using namespace std;
#define ull unsigned long long
#define ll long long
#define R register ll
#define pause (for(R i=1;i<=10000000000;++i))
#define In freopen("NOIPAK++.in","r",stdin)
#define Out freopen("out.out","w",stdout)
namespace Fread {
static char B[<<],*S=B,*D=B;
#ifndef JACK
#define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++)
#endif
inline ll g() {
    R ret=,fix=; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-:fix;
    if(ch==EOF) return EOF; do ret=ret*+(ch^); while(isdigit(ch=getchar())); return ret*fix;
} inline bool isempty(const char& ch) {return (ch<=||ch>=);}
inline void gs(char* s) {
    register char ch; while(isempty(ch=getchar()));
    do *s++=ch; while(!isempty(ch=getchar()));
}
} using Fread::g; using Fread::gs;

namespace Luitaryi {
const int N=3e++;
int n,m,k,lim,ct;
struct node {int fir,rk; ll w;}a[N],tmp[N];
#define fir(u) a[u].fir
int vr[N],nxt[N],LL[N],RR[N],W[N],ans[N];
ll c[N<<];
inline void adde(int u,int v) {vr[++ct]=v,nxt[ct]=fir(u),fir(u)=ct;}
inline int lbt(int x) {return x&-x;}
inline void add(int pos,int inc) {for(;pos<=lim;pos+=lbt(pos)) c[pos]+=inc;}
inline ll query(int pos) { R ret=;
    for(;pos;pos-=lbt(pos)) ret+=c[pos]; return ret;
}
inline void solve(int l,int r,int s,int t) {
    if(l==r) {for(R i=s;i<=t;++i) ans[a[i].rk]=l; return ;}
    R md=l+r>>,f=s-,b=t+;
    for(R i=l;i<=md;++i) add(LL[i],W[i]),add(RR[i]+,-W[i]);
    for(R u=s;u<=t;++u) { R cnt=;
        for(R i=fir(u),lim=a[u].w;i&&cnt<lim;i=nxt[i]) { R v=vr[i];
            cnt+=query(v)+query(v+m);
        } if(cnt>=a[u].w) tmp[++f]=a[u];
        else tmp[--b]=a[u],tmp[b].w-=cnt;
    } for(R i=l;i<=md;++i) add(LL[i],-W[i]),add(RR[i]+,W[i]);
    for(R i=s;i<=f;++i) a[i]=tmp[i];
    for(R i=t;i>=b;--i) a[i]=tmp[i];
    solve(l,md,s,f); solve(md+,r,b,t);
}
inline void main() {
    n=g(),m=g(),lim=m<<;
    for(R i=,x;i<=m;++i) x=g(),adde(x,i);
    for(R i=;i<=n;++i) a[i].w=g(),a[i].rk=i;
    k=g(); for(R i=;i<=k;++i) LL[i]=g(),RR[i]=g(),W[i]=g(),LL[i]>RR[i]?RR[i]+=m:;
    solve(,k+,,n); for(R i=;i<=n;++i) ans[i]==k+?printf("NIE\n"):printf("%d\n",ans[i]);
}
}
signed main() {
    Luitaryi::main();
    return ;
}

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2019.07.14