线段树 csu1555 Inversion Sequence

传送门：​​点击打开链接​​

题意：告诉你每个数字的逆序数，求原序列

思路：从1开始考虑，利用线段树记录空白位置的数量，然后利用线段树维护插入位置经行插入

#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck printf("fuck")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;

const int MX = 10000 + 5;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define root 1,n,1

int S[MX << 2], A[MX], B[MX];

void push_up(int rt) {
    S[rt] = S[rt << 1] + S[rt << 1 | 1];
}

void build(int l, int r, int rt) {
    if(l == r) {
        S[rt] = 1;
        return;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    push_up(rt);
}

int query(int x, int l, int r, int rt) {
    if(l == r) {
        S[rt]--;
        return l;
    }

    int m = (l + r) >> 1, ret;
    if(x <= S[rt << 1]) ret = query(x, lson);
    else ret = query(x - S[rt << 1], rson);
    push_up(rt);
    return ret;
}

int main() {
    int n; //FIN;
    while(~scanf("%d", &n)) {
        build(root);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &A[i]);
        }

        bool sign = false;
        for(int i = 1; i <= n; i++) {
            if(S[1] < A[i] + 1) {
                sign = true;
                break;
            }
            int p = query(A[i] + 1, root);
            B[p] = i;
        }

        if(sign) printf("No solution\n");
        else for(int i = 1; i <= n; i++) {
                printf("%d%c", B[i], i == n ? '\n' : ' ');
            }
    }
    return 0;
}