uva 11525 - Permutation(线段树)

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题目链接：uva 11525 - Permutation

题目大意：给定n和k，n给定的方式为k个si,根据公式计算出n，求一个由1~k组成的长度为k的序列的第n个排序

解题思路：根据公式的性质，等于对于每个位置找当前状态下第si小的数。线段树子节点均为1，维护和，查询时传入参数查找即可。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 50005;
#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)+1)

struct Node {
    int l, r, val;
    void set (int l, int r, int val) {
        this->l = l;
        this->r = r;
        this->val = val;
    }
}node[maxn * 4];

void pushup(int u) {
    node[u].set(node[lson(u)].l, node[rson(u)].r, node[lson(u)].val + node[rson(u)].val);
}
void build_segTree (int u, int l, int r) {
    if (l == r) {
        node[u].set(l, r, 1);
        return ;
    }

    int mid = (l + r) / 2;
    build_segTree(lson(u), l, mid);
    build_segTree(rson(u), mid + 1, r);
    pushup(u);
}

int query_segTree (int u, int val) {
    if (node[u].l == node[u].r && val == 1)
        return node[u].l;

    if (val > node[lson(u)].val)
        return query_segTree(rson(u), val - node[lson(u)].val);
    else
        return query_segTree(lson(u), val);
}

void insert_segTree (int u, int x, int val) {
    if (node[u].l == x && node[u].r == x) {
        node[u].val += val;
        return ;
    }

    int mid = (node[u].l + node[u].r) / 2;
    if (x <= mid)
        insert_segTree(lson(u), x, val);
    else
        insert_segTree(rson(u), x, val);
    pushup(u);
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        int n, x;
        scanf("%d", &n);
        build_segTree(1, 1, n);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &x);
            x = query_segTree(1, x+1);
            insert_segTree(1, x, -1);
            printf("%d%c", x, i == n ? ‘\n‘ : ‘ ‘);
        }
    }
    return 0;
}

uva 11525 - Permutation(线段树)

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/38820105