P6035 Ryoku 的逆序对 （segment tree）

题目链接

题解：

 copy from: https://www.luogu.com.cn/blog/ryoku/ryoku-di-xin-nian-huan-le-sai-ti-xie

segment tree, seg[i]为区间元素个数，每次从剩下的数中找第（b[i]+1）小的数，然后去掉次数（seg[i]=0）.

Code:

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

const int maxn=1000000+10;
const int mod=1E9+7;
int seg[maxn<<2];
int n;
int b[maxn];

void build(int idx, int left, int right){
    if(left==right){
        seg[idx]=1;
        return;
    }

    int mid=(left+right)>>1;
    build(idx<<1,left,mid);
    build((idx<<1)+1,mid+1,right);
    seg[idx]=seg[idx<<1]+seg[(idx<<1)+1];
}

int findKth(int idx, int left, int right, int k){
    if(left==right) return left;

    int size=seg[idx<<1];
    int mid=(left+right)>>1;
    if(k<=size) return findKth(idx<<1,left,mid,k);
    else return findKth((idx<<1)+1,mid+1,right,k-size);
}

void modify(int idx, int left, int right, int x){
    if(left==right){
        seg[idx]=0;
        return;
    }
    int mid=(left+right)>>1;
    if(x<=mid) modify(idx<<1,left,mid,x);
    else modify((idx<<1)+1,mid+1,right,x);
    seg[idx]=seg[idx<<1]+seg[(idx<<1)+1];
}

int main(){
    scanf("%d", &n);
    for(int i=1;i<=n;++i){
        scanf("%d", b+i);
    }
    LL ans=1;
    for(int i=1;i<=n;++i){
        if(b[i]>n-i){
            printf("0");
            return 0;
        }
        if(b[i]==-1){
            ans=(ans*(n-i+1))%mod;
            b[i]=0;
        }
    }
    printf("%d
", ans);
    build(1,1,n);
    for(int i=1;i<=n;++i){
        int x=findKth(1,1,n,b[i]+1);
        printf("%d ",x);
        modify(1,1,n,x);
    }
    return 0;
}