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=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 300010;
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+1; p<=n; p+=p&(-p)) sum[p] -= v;
        if (l > r) for (int p=1; p<=n; p+=p&(-p)) sum[p] += v;
    }
    inline LL query(int p) { // LL
        LL ans = 0;
        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] = 0;
        for (int p=r+1; p<=n&&sum[p]; p+=p&(-p)) sum[p] = 0;
        if (l > r) for (int p=1; p<=n&&sum[p]; p+=p&(-p)) sum[p] = 0;
    }
}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) / 2,p1 = 0,p2 = 0;
    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 = 0;
        for (int j=0,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=1; i<=p1; ++i) A[Head+i-1] = B1[i];
    for (int i=1; i<=p2; ++i) A[Head+p1+i-1] = B2[i];
    solve(Head,Head+p1-1,L,M);
    solve(Head+p1,Tail,M+1,R);
}

int main() {
    int n,m,k;
    n = read();
    bit.n = m = read();
    for (int i=1; i<=m; ++i) v[read()].push_back(i); // 第read()个国家的空间站i
    for (int i=1; i<=n; ++i) T[i] = read(),A[i] = i;
    k = read();
    for (int i=1; i<=k; ++i) 
        OP[i].l = read(),OP[i].r = read(),OP[i].val = read();
    OP[++k] = (OPT){m+1,m+1,0}; // 特判不行的情况 
    
    solve(1,n,1,k);
    for (int i=1; i<=n; ++i) printf(Ans[i]==k?"NIE
":"%d
",Ans[i]);
    return 0;
}