P3527 [POI2011]MET-Meteors

P3527 [POI2011]MET-Meteors

链接

　　整体二分！

代码

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long LL;

 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for (;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }

 const int N = ;
 int Ans[N],A[N],B1[N],B2[N],T[N];
 vector <int> v[N];
 struct OPT{
     int l,r,val;
 }OP[N];

 struct BIT{
     int n;
     LL sum[N];
     inline void update(int l,int r,int v) {
         for (int p=l; p<=n; p+=p&(-p)) sum[p] += v;
         for (int p=r+; p<=n; p+=p&(-p)) sum[p] -= v;
         if (l > r) for (int p=; p<=n; p+=p&(-p)) sum[p] += v;
     }
     inline LL query(int p) { // LL
         LL ans = ;
         for (; p; p-=p&(-p)) ans += sum[p];
         return ans;
     }
     inline void clear(int l,int r) {
         for (int p=l; p<=n&&sum[p]; p+=p&(-p)) sum[p] = ;
         for (int p=r+; p<=n&&sum[p]; p+=p&(-p)) sum[p] = ;
         if (l > r) for (int p=; p<=n&&sum[p]; p+=p&(-p)) sum[p] = ;
     }
 }bit;

 void solve(int Head,int Tail,int L,int R) { // 询问的队列（即国家）[Head,Tail]，答案[L,R]次之间
     if (Head > Tail) return;
     if (L == R) {
         for (int i=Head; i<=Tail; ++i) Ans[A[i]] = L;
         return;
     }
     int M = (L + R) / ,p1 = ,p2 = ;
     for (int i=L; i<=M; ++i)
         bit.update(OP[i].l,OP[i].r,OP[i].val);
     for (int i=Head; i<=Tail; ++i) {
         int now = A[i]; // 国家
         LL sum = ;
         for (int j=,lim=v[now].size(); j<lim; ++j)
             if ((sum += bit.query(v[now][j])) >= T[now]) break;;
         if (sum >= T[now]) B1[++p1] = A[i];
         else T[now] -= sum,B2[++p2] = A[i];
     }
     for (int i=L; i<=M; ++i) bit.clear(OP[i].l,OP[i].r);
     for (int i=; i<=p1; ++i) A[Head+i-] = B1[i];
     for (int i=; i<=p2; ++i) A[Head+p1+i-] = B2[i];
     solve(Head,Head+p1-,L,M);
     solve(Head+p1,Tail,M+,R);
 }

 int main() {
     int n,m,k;
     n = read();
     bit.n = m = read();
     for (int i=; i<=m; ++i) v[read()].push_back(i); // 第read()个国家的空间站i
     for (int i=; i<=n; ++i) T[i] = read(),A[i] = i;
     k = read();
     for (int i=; i<=k; ++i)
         OP[i].l = read(),OP[i].r = read(),OP[i].val = read();
     OP[++k] = (OPT){m+,m+,}; // 特判不行的情况 

     solve(,n,,k);
     for (int i=; i<=n; ++i) printf(Ans[i]==k?"NIE\n":"%d\n",Ans[i]);
     return ;
 }