【模拟】NCPC 2014 E ceremony
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题目链接:
http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1791
题目大意:
N栋大楼(N<=100000),拆除的时候有两种选择的操作,可以拆掉一整栋,或者把当前所有大楼的第x层拆掉(高度小于x的没用),并且x以上的所有楼层高度-1.(相当于所有高于x的大楼抽掉一层)
问把N栋大楼拆掉至少要几次拆除操作。
题目思路:
【模拟】
先将大楼高度排序,之后考虑分界线,枚举I,I之后的大楼都是被单个拆除,而I和I之前的大楼都是被一层一层拆除。ans=min(n-i+h[i])
////by coolxxx//#include<bits/stdc++.h>#include<iostream>#include<algorithm>#include<string>#include<iomanip>#include<map>#include<stack>#include<queue>#include<set>#include<bitset>#include<memory.h>#include<time.h>#include<stdio.h>#include<stdlib.h>#include<string.h>//#include<stdbool.h>#include<math.h>#define min(a,b) ((a)<(b)?(a):(b))#define max(a,b) ((a)>(b)?(a):(b))#define abs(a) ((a)>0?(a):(-(a)))#define lowbit(a) (a&(-a))#define sqr(a) ((a)*(a))#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))#define mem(a,b) memset(a,b,sizeof(a))#define eps (1e-8)#define J 10#define mod 1000000007#define MAX 0x7f7f7f7f#define PI 3.14159265358979323#define N 100004using namespace std;typedef long long LL;int cas,cass;int n,m,lll,ans;int h[N];bool cmp(int a,int b){return a<b;}int main(){#ifndef ONLINE_JUDGE//freopen("1.txt","r",stdin);//freopen("2.txt","w",stdout);#endifint i,j,k;//for(scanf("%d",&cass);cass;cass--)//for(scanf("%d",&cas),cass=1;cass<=cas;cass++)//while(~scanf("%s",s+1))while(~scanf("%d",&n)){for(i=1;i<=n;i++)scanf("%d",&h[i]);sort(h+1,h+1+n,cmp);ans=min(n,h[n]);for(i=1;i<n;i++)ans=min(ans,n-i+h[i]);printf("%d\n",ans);}return 0;}/*////*/
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