zoukankan      html  css  js  c++  java
  • codeforces489A——SwapSort

    Description

    In this problem your goal is to sort an array consisting of n integers in at most n swaps. For the given array find the sequence of swaps that makes the array sorted in the non-descending order. Swaps are performed consecutively, one after another.

    Note that in this problem you do not have to minimize the number of swaps — your task is to find any sequence that is no longer than n.

    Input

    The first line of the input contains integer n (1 ≤ n ≤ 3000) — the number of array elements. The second line contains elements of array: a0, a1, ..., an - 1 ( - 109 ≤ ai ≤ 109), where ai is the i-th element of the array. The elements are numerated from 0 to n - 1from left to right. Some integers may appear in the array more than once.

    Output

    In the first line print k (0 ≤ k ≤ n) — the number of swaps. Next k lines must contain the descriptions of the k swaps, one per line. Each swap should be printed as a pair of integers ij (0 ≤ i, j ≤ n - 1), representing the swap of elements ai and aj. You can print indices in the pairs in any order. The swaps are performed in the order they appear in the output, from the first to the last. It is allowed to print i = jand swap the same pair of elements multiple times.

    If there are multiple answers, print any of them. It is guaranteed that at least one answer exists.

    Sample Input

    Input
    5
    5 2 5 1 4
    Output
    2
    0 3
    4 2
    Input
    6
    10 20 20 40 60 60
    Output
    0
    Input
    2
    101 100
    Output
    1
    0 1
    大意:让你求出交换次数以及每次交换的元素
    不知道为什么用归并做WA,用vector遍历每一个A了。。
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<vector>
    using namespace std;
    vector<pair <int ,int> >ans;
    int a[3000],n;
    int main(){
        scanf("%d",&n);
        for(int i = 0; i < n ; i++)
        scanf("%d", &a[i]);
        for(int i = n - 1; i >= 0; i--){
                int p = max_element(a,a+i+1)-a;
                swap(a[p],a[i]);
                ans.push_back(make_pair(i,p));
        }
        printf("%d
    ",ans.size());
        for(int i = 0; i < ans.size();i++)
            printf("%d %d
    ",ans[i].first,ans[i].second);
       return 0;
    }
    View Code

     学到了vector的max_elememt的用法,如max_element(a,a+i+1)就是求其之间的最大元素的下标!注意是下标还要减去a,用容器的size比较简单

     
  • 相关阅读:
    JavaFx在macOS下的文字渲染Bug
    Java多接口同名方法的冲突
    旧技术的惯性
    一点思考(1)
    slisp:编译到JVM平台上的lisp方言
    Arcee:又一个 Parser Generator 轮子
    使用Java实现一门简单的动态语言
    Hello World!
    [NOI2008]奥运物流
    [IOI2005]Riv河流
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4369916.html
Copyright © 2011-2022 走看看