Codeforces Round #480 (Div. 2)980C Posterized+分组类贪心
传送门:http://codeforces.com/contest/980/problem/C
题意:
给定n个数字,每个数在0~256间,现在给至多连续k的数分为一组,给出字典序最小的答案。
思路:
贪心,对于每一个a[i],查找$max(0,a[i] - k + 1)$到 a[ i ] 中符合$vis[ t ] ==t || vis[ t ] = -1$的最小数字 T ,把这个T到 a[ i ]间的更新为T;
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> using namespace std; const int maxn = 1e5+9; const int inf = 0x3f3f3f3f; int n,k; int a[maxn],vis[maxn]; int d[maxn]; int main(){ scanf("%d%d", &n,&k); for(int i=1; i<=n; i++) scanf("%d", &a[i]); memset(vis,-1,sizeof(vis)); for(int i=1;i <= n; i++) { if(vis[a[i]]!=-1) printf("%d%c",vis[a[i]]," \n"[i==n]); else { int j; int tot = 0; for(j = max(0,a[i] - k + 1); j<=a[i]; j++) { if(vis[j] == -1||vis[j] == j) { break; } } for(int g = j;g<=a[i];g++) { vis[g] = j; } printf("%d%c",vis[a[i]]," \n"[i==n]); } } return 0; }
skr