P3527 [POI2011]MET-Meteors
P3527 [POI2011]MET-Meteors
整体二分！
代码
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 5 inline int read() { 6 int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; 7 for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f; 8 } 9 10 const int N = 300010; 11 int Ans[N],A[N],B1[N],B2[N],T[N]; 12 vector <int> v[N]; 13 struct OPT{ 14 int l,r,val; 15 }OP[N]; 16 17 struct BIT{ 18 int n; 19 LL sum[N]; 20 inline void update(int l,int r,int v) { 21 for (int p=l; p<=n; p+=p&(-p)) sum[p] += v; 22 for (int p=r+1; p<=n; p+=p&(-p)) sum[p] -= v; 23 if (l > r) for (int p=1; p<=n; p+=p&(-p)) sum[p] += v; 24 } 25 inline LL query(int p) { // LL 26 LL ans = 0; 27 for (; p; p-=p&(-p)) ans += sum[p]; 28 return ans; 29 } 30 inline void clear(int l,int r) { 31 for (int p=l; p<=n&&sum[p]; p+=p&(-p)) sum[p] = 0; 32 for (int p=r+1; p<=n&&sum[p]; p+=p&(-p)) sum[p] = 0; 33 if (l > r) for (int p=1; p<=n&&sum[p]; p+=p&(-p)) sum[p] = 0; 34 } 35 }bit; 36 37 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 38 if (Head > Tail) return; 39 if (L == R) { 40 for (int i=Head; i<=Tail; ++i) Ans[A[i]] = L; 41 return; 42 } 43 int M = (L + R) / 2,p1 = 0,p2 = 0; 44 for (int i=L; i<=M; ++i) 45 bit.update(OP[i].l,OP[i].r,OP[i].val); 46 for (int i=Head; i<=Tail; ++i) { 47 int now = A[i]; // 国家 48 LL sum = 0; 49 for (int j=0,lim=v[now].size(); j<lim; ++j) 50 if ((sum += bit.query(v[now][j])) >= T[now]) break;; 51 if (sum >= T[now]) B1[++p1] = A[i]; 52 else T[now] -= sum,B2[++p2] = A[i]; 53 } 54 for (int i=L; i<=M; ++i) bit.clear(OP[i].l,OP[i].r); 55 for (int i=1; i<=p1; ++i) A[Head+i-1] = B1[i]; 56 for (int i=1; i<=p2; ++i) A[Head+p1+i-1] = B2[i]; 57 solve(Head,Head+p1-1,L,M); 58 solve(Head+p1,Tail,M+1,R); 59 } 60 61 int main() { 62 int n,m,k; 63 n = read(); 64 bit.n = m = read(); 65 for (int i=1; i<=m; ++i) v[read()].push_back(i); // 第read()个国家的空间站i 66 for (int i=1; i<=n; ++i) T[i] = read(),A[i] = i; 67 k = read(); 68 for (int i=1; i<=k; ++i) 69 OP[i].l = read(),OP[i].r = read(),OP[i].val = read(); 70 OP[++k] = (OPT){m+1,m+1,0}; // 特判不行的情况 71 72 solve(1,n,1,k); 73 for (int i=1; i<=n; ++i) printf(Ans[i]==k?"NIE\n":"%d\n",Ans[i]); 74 return 0; 75 }
