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Codeforces 489A SwapSort

这题第一次看的时候以为是区间替换,后来发现看错了,只是单纯的元素替换。

解题思路:

  先对输入的序列加个数组排个序

  遍历下来,如果和排序后的结果当前元素不同,设当前位置为 i, 则往下面找,设查找位置为j

  使得满足 a[j] == b[i] && a[j] != b[j]

  一次遍历即可。易得证

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler#include <stdio.h>#include <iostream>#include <cstring>#include <cmath>#include <stack>#include <queue>#include <vector>#include <algorithm>#define ll long long#define Max(a,b) (((a) > (b)) ? (a) : (b))#define Min(a,b) (((a) < (b)) ? (a) : (b))#define Abs(x) (((x) > 0) ? (x) : (-(x)))using namespace std;const int INF = 0x3f3f3f3f;int a[3111], b[3111], n, counti;int ans[3111][2];int main(){    int i, j, t, m;    while(EOF != scanf("%d",&n)){        counti = 0;        for(i = 1; i <= n; ++i){            scanf("%d",&a[i]);            b[i] = a[i];        }        sort(b + 1, b + 1 + n);        for(i = 1; i <= n; ++i){            if(a[i] != b[i]){                for(j = i + 1; j <= n; ++j){                    if(a[j] == b[i] && a[j] != b[j]){                        swap(a[j], a[i]);                        ++counti;                        ans[counti][0] = j;                        ans[counti][1] = i;                        break;                    }                }            }        }        printf("%d\n",counti);        for(i = 1; i <= counti; ++i){            printf("%d %d\n",ans[i][1] - 1, ans[i][0] - 1);        }    }    return 0;}

 

Codeforces 489A SwapSort