p3527 [POI2011]MET-Meteors

传送门

分析

一个整体二分的经典题

每次二分得到两个答案区间，然后把现在的国家分别看属于哪个答案区间，递归求解即可

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
int n,m,k,now,t1[300100],t2[300100],d[300100];
int sum[300100],id[300100],L[300100],R[300100],w[300100],Ans[300100];
vector<int>wh[300100];
inline int lb(int x){return x&(-x);}
inline int q(int x){int res=0;while(x)res+=d[x],x-=lb(x);return res;}
inline void add(int x,int s){while(x<=m)d[x]+=s,x+=lb(x);}
inline bool ck(int x){
    int res=0;
    for(int i=0;i<wh[x].size();i++){
      res+=q(wh[x][i]);
      if(res>=sum[x])return 1;
    }
    return 0;
}
inline void update(int x,int s){
    add(L[x],s),add(R[x]+1,-s);
    if(R[x]<L[x])add(1,s);
}
inline void work(int x,int y,int le,int ri){
    if(x>y||le>ri)return;
    int mid=(le+ri)>>1,l1=0,l2=0;
    if(le==ri){
      for(int i=x;i<=y;i++)Ans[id[i]]=le;
      return;
    }
    while(now<mid)now++,update(now,w[now]);
    while(now>mid)update(now,-w[now]),now--;
    for(int i=x;i<=y;i++)
      if(ck(id[i]))t1[++l1]=id[i];
        else t2[++l2]=id[i];
    for(int i=1;i<=l1;i++)id[x+i-1]=t1[i];
    for(int i=1;i<=l2;i++)id[x+l1+i-1]=t2[i];
    if(l1)work(x,x+l1-1,le,mid);
    if(l2)work(x+l1,y,mid+1,ri);
}
int main(){
    int i,j,x;
    scanf("%d%d",&n,&m);
    for(i=1;i<=m;i++){
      scanf("%d",&x);
      wh[x].push_back(i);
    }
    for(i=1;i<=n;i++)scanf("%d",&sum[i]),id[i]=i;
    scanf("%d",&k);k++;
    for(i=1;i<k;i++)scanf("%d%d%d",&L[i],&R[i],&w[i]);
    L[k]=1,R[k]=m,w[k]=1e9+7;
    work(1,m,1,k);
    for(i=1;i<=n;i++)
      if(Ans[i]==k)puts("NIE");
        else printf("%d\n",Ans[i]);
    return 0;
}