Codeforces Round #436 (Div. 2) Make a Permutation!
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题目链接:http://codeforces.com/contest/864/problem/D
题意:给你一个序列,要使不能在序列中出现相同的数字,问你最少要改多少个数字,要求字典序最小
思路:这个题看起来很容易,其实实现起来还是有点麻烦的,首先是找到出现重复的数字,并记录它出现的次数,然后就是用一个数组来从小到大来存可以加进来的数,最后是改数,如果比加的数大就直接换了,如果比加的数小那么第一个出现的数就不改,并且后面的出现的所有这个数都要改了,从而保证只有一个不改变,而顺序是从小到大,具体看代码。
#include <bits/stdc++.h>using namespace std;typedef long long LL;typedef pair<LL, LL> P;const int maxn = 2e5 + 5;const int mod = 1e8 + 7;int n;struct node { //一开始写复杂了,所以用了个结构体,其实不用的,2333,就当是个普通数组啦,懒得改了 int value;}a[maxn];int num[maxn];int res[maxn];bool vis[maxn];int main() { //freopen ("in.txt", "r", stdin); while (~scanf ("%d",&n)){ memset(num,0,sizeof(num)); memset(res,0,sizeof(res)); memset(vis,0,sizeof(vis)); int Max=0; for (int i=1;i<=n;i++){ scanf ("%d",&a[i].value); if (a[i].value>Max) Max=a[i].value; num[a[i].value]++; } for (int i=1;i<=n;i++){ if (num[a[i].value]==1) { num[a[i].value]=-1; } } int cnt=0; int s=0,ss=0; for (int i=1;i<=n;i++){ if (num[i]==0) { res[cnt++]=i; } else if (num[i]>=2) { s++; ss+=num[i]; } } while (cnt<ss-s){ res[cnt++]=++Max; } int x=0; for (int i=1;i<=n;i++){ if (num[a[i].value]>=2) { if (a[i].value>res[x]){ num[a[i].value]--; a[i].value=res[x]; x++; } else { if (vis[a[i].value]==0){ vis[a[i].value]=1; continue; } num[a[i].value]--; a[i].value=res[x]; x++; } } } printf ("%d\n",ss-s); for (int i=1;i<=n;i++){ printf ("%d ",a[i].value); } printf ("\n"); } return 0;}
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