2017-2018 ACM-ICPC, NEERC, Southern Subregional Contest, qualification stage (Online Mirror, ACM-ICPC Rules, Teams Preferred) H. Load Testing

题意：给出一个数列，问修改差值最小，使得整个数列为严格递增，严格递减的数列，连续2个数字不能相同

思路：正的，反的跑一遍，记录当前这个如果为峰顶，左边要改多少，右边要改多少，

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int N=200005;
 5 ll a[N];
 6 ll l[N],r[N];
 7 ll c[N];
 8 
 9 int main(){
10     int n;
11     cin>>n;
12     ll Max=0;
13     int sss=0;
14     for(int i=1;i<=n;i++){
15         scanf("%lld",&a[i]);
16         Max=max(Max,a[i]);
17     }
18     ll M=0;
19     for(int i=1;i<=n;i++){
20         if(a[i]<=M)
21         {
22             l[i]=l[i-1]+(M+1-a[i]);
23             M++;
24         }
25         else {
26             l[i]=l[i-1];
27         }
28         M=max(M,a[i]);
29         c[i]=M;
30       //  cout<<c[i]<<" ";
31     }
32     //cout<<endl;
33      M=0;
34     for(int i=n;i>=1;i--){
35         if(a[i]<=M){
36             r[i]=r[i+1]+M+1-a[i];
37             M++;
38         }
39         else r[i]=r[i+1];
40         M=max(M,a[i]);
41         c[i]=max(c[i],M);
42     }
43     ll Min=1e18;
44     Max++;
45     for(int i=1;i<=n;i++){
46         ll sum=c[i]-a[i];
47         sum+=l[i-1]+r[i+1];
48        // cout<<c[i]<<endl;
49         Min=min(Min,sum);
50     }
51     cout<<Min<<endl;
52 }

 

posted on 2017-09-20 18:53  hhhhx  阅读(173)  评论(0)  编辑  收藏  举报

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