P2617 Dynamic Rankings

P2617 Dynamic Rankings

链接

分析：

　　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 = 10010;
struct OPT{
    int l,r,k,ty,id;
}A[N*3],B1[N*3],B2[N*3];
int ans[N],last[N],n,m;

struct BIT{
    int sum[N];
    inline void update(int p,int v) {
        for (; p<=n; p+=p&(-p)) sum[p] += v;
    }
    inline int query(int p) {
        int ans = 0;
        for (; p; p-=p&(-p)) ans += sum[p];
        return ans;
    }
}bit;

void solve(int Head,int Tail,int L,int R) {
    if (Head > Tail) return ;
    if (L == R) {
        for (int i=Head; i<=Tail; ++i) 
            if (A[i].ty == 2) ans[A[i].id] = L;
        return; 
    }
    int M = (L + R) >> 1,p1 = 0,p2 = 0;
    for (int i=Head; i<=Tail; ++i) {
        if (A[i].ty <= 1) {
            if (A[i].k <= M) {B1[++p1] = A[i];bit.update(A[i].l,A[i].ty?-1:1);}
            else B2[++p2] = A[i];
        }
        else {
            int tmp = bit.query(A[i].r)-bit.query(A[i].l-1);
            if (tmp >= A[i].k) B1[++p1] = A[i];
            else A[i].k -= tmp,B2[++p2] = A[i];
        }
    }
    for (int i=Head; i<=Tail; ++i) 
        if (A[i].ty <= 1 && A[i].k <= M) bit.update(A[i].l,A[i].ty?1:-1);
    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() {
    n = read(),m = read();
    int QueIndex = 0,Idx = 0,Mx = 0,Mn = 1e9;
    for (int i=1; i<=n; ++i) {
        int x = read();
        A[++Idx] = (OPT){i,i,x,0,0};
        last[i] = x; 
        Mx = max(Mx,x);Mn = min(Mn,x);
    }
    char opt[5];
    for (int i=1; i<=m; ++i) {
        scanf("%s",opt);
        if (opt[0] == 'Q') {
            int a = read(),b = read(),c = read();
            A[++Idx] = (OPT){a,b,c,2,++QueIndex};
        }
        else {
            int p = read(),x = read();
            A[++Idx] = (OPT){p,p,last[p],1,0};
            A[++Idx] = (OPT){p,p,last[p]=x,0,0};
            Mx = max(Mx,x);Mn = min(Mn,x);
        }
    }
    solve(1,Idx,Mn,Mx);
    for (int i=1; i<=QueIndex; ++i) printf("%d
",ans[i]);
    return 0;
}