UVa 11525 - Permutation （线段树 树状数组 康托展开式）

UVA - 11525

Permutation

Time Limit: 3000MSMemory Limit: Unknown64bit IO Format: %lld & %llu

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Description

Problem F
Permutation
Input: Standard Input

Output: Standard Output

 

Given N and K find the N’th permutation of the integers from 1 to K when those permutations are lexicographically ordered.  N starts from 0. Since N is very large N will be represented by a sequence of K non-negative integers S₁, S₂ ,…, S_k. From this sequence of integers N can be calculated with the following expression.

 

 

Input

First line of the input contains T(≤10) the number of test cases. Each of these test cases consists of 2 lines. First line contains a integer K(1≤K≤50000). Next line contains K integers S₁, S₂ ,…, S_k.(0≤S_i≤K-i).

 

Output

For each test case output contains N’th permutation of the integers from 1 to K. These K integers should be separated by a single space.

 

Sample Input                            Output for Sample Input

4 

3 

2 1 0 

3 

1 0 0 

4 

2 1 1 0 

4 

1 2 1 0 

 

3 2 1 

2 1 3 

3 2 4 1 

2 4 3 1 

 

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Problemsetter: Abdullah al Mahmud

Special Thanks: Manzurur Rahman Khan

 

Source

Root :: Prominent Problemsetters :: Abdullah-al-Mahmud (Satej)
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: More Advanced Topics :: Problem Decomposition :: Two Components - Mixed with Efficient Data Structure
Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 3. Data Structures :: Maintaining Interval Data ::Exercises: Beginner

Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Data Structures and Libraries :: Data Structures with Our-Own Libraries :: Tree-related Data Structures

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题意：

给定整数n和k，输出1-k的所有排列中，按照字典序从小到大排序后的第n个（编号0开始）。

由于n通过下面的方式给出

n =  

康拓展开式逆运算

点我~

线段树版本

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <map>#include <cmath>#include <queue>#include <set>using namespace std;//#define WIN#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#define oform1 "%I64d"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#define oform1 "%lld"#endif#define S64I(a) scanf(iform, &(a))#define P64I(a) printf(oform, (a))#define P64I1(a) printf(oform1, (a))#define REP(i, n) for(int (i)=0; (i)<n; (i)++)#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)#define lowbit(x) (x&-x)const int INF = 0x3f3f3f3f;const double eps = 1e-9;const double PI = (4.0*atan(1.0));const int maxn = 50000 + 20;const int maxo = maxn * 4;int S[maxn];int ans[maxn];int sumv[maxo];void pushUp(int o) {    sumv[o] = sumv[o<<1] + sumv[o<<1|1];}void build(int o, int L, int R) {    if(L == R) {        sumv[o] = 1;        return ;    }    int M = L + (R-L) / 2;    build(o<<1, L, M);    build(o<<1|1, M+1, R);    pushUp(o);}int qx, qv;int update(int o, int L, int R) {    if(L == R) {        sumv[o] = qv;        return L;    }    int M = L + (R-L) / 2;    int ret;    if(sumv[o<<1] >= qx) ret = update(o<<1, L, M);    else {        qx -= sumv[o<<1];        ret = update(o<<1|1, M+1, R);    }    pushUp(o);    return ret;}int main() {    int T;    scanf("%d", &T);    while(T--) {        int n;        scanf("%d", &n);        for(int i=0; i<n; i++) scanf("%d", &S[i]);        build(1, 1, n);        for(int i=0; i<n; i++) {            qx = S[i]+1, qv = 0;            ans[i] = update(1, 1, n);        }        for(int i=0; i<n; i++) {            if(i) putchar(' ');            printf("%d", ans[i]);        }        putchar('\n');    }    return 0;}

树状数组+二分

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <map>#include <cmath>#include <queue>#include <set>using namespace std;//#define WIN#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#define oform1 "%I64d"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#define oform1 "%lld"#endif#define S64I(a) scanf(iform, &(a))#define P64I(a) printf(oform, (a))#define P64I1(a) printf(oform1, (a))#define REP(i, n) for(int (i)=0; (i)<n; (i)++)#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)#define lowbit(x) (x&-x)const int INF = 0x3f3f3f3f;const double eps = 1e-9;const double PI = (4.0*atan(1.0));const int maxn = 50000 + 20;int S[maxn];int ans[maxn];int C[maxn];int vis[maxn];int n;void add(int x, int v) {    while(x <= n) {        C[x] += v;        x += lowbit(x);    }}int sum(int x) {    int ret = 0;    while(x) {        ret += C[x];        x -= lowbit(x);    }    return ret;}int main() {    int T;    scanf("%d", &T);    while(T--) {        scanf("%d", &n);        for(int i=0; i<n; i++) scanf("%d", &S[i]);        memset(C, 0, sizeof(C));        memset(vis, 0, sizeof(vis));        for(int i=1; i<=n; i++) add(i, 1);        for(int i=0; i<n; i++) {            int L = 1, R = n;            int ansi = 1;            while(L <= R) {                int M = L + (R-L) / 2;                int tot = sum(M);                if(tot == S[i]+1 && !vis[M]) {                    ansi = M;                    break;                }                if(tot < S[i]+1) L = M + 1;                else R = M - 1;            }            vis[ansi] = 1;            add(ansi, -1);            ans[i] = ansi;        }        for(int i=0; i<n; i++) {            if(i) putchar(' ');            printf("%d", ans[i]);        }        putchar('\n');    }    return 0;}