[codeforces1221D] Make The Fence Great Again dp

题目：

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have a fence consisting of

Unfortunately, it is possible that now your fence is not great. But you can change it! You can increase the length of the

Calculate the minimum number of rubles you have to spend to make the fence great again!

You have to answer

Input

The first line contains one integer

The first line of each query contains one integers

The following

It is guaranteed that sum of all

It is guaranteed that answer to each query will not exceed

Output

For each query print one integer — the minimum number of rubles you have to spend to make the fence great.

Example

input

Copy

3
3
2 4
2 1
3 5
3
2 3
2 10
2 6
4
1 7
3 3
2 6
1000000000 2

output

Copy

2
9
0

Note

In the first query you have to increase the length of second board by

In the second query you have to increase the length of first board by

In the third query the fence is great initially, so you don't need to spend rubles.

题意：给一列数ai，使相邻的数不同，可以花费bi代价使ai增加1，求最小代价

题解：

参考的题解：https://www.cnblogs.com/bianjunting/p/11556876.html
比赛的时候没做出来，没想到一个栏杆最多只需要增加两次。
因为如果两个相邻的数a[i] 和 a[i-1]相同，选择其中一个+1，比如a[i],这时候如果a[i]==a[i+1],a[i]再+1就能和a[i+1]不同。
线性dp f[i][k]表示把第i个数增加k次，使i和之前的数满足条件的最小代价。k=0,1,2。
if(a[i]+k!=a[i-1]+j)
f[i][k]=min(f[i-1][j]+k*b[i],f[i][k]);

ac代码

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int const maxn=300010;
 4 long long  a[maxn],n,b[maxn],f[maxn][4];
 5 inline long long  get_num(){
 6     long long  num=0;
 7     char ch=getchar();
 8     while(ch<'0'||ch>'9')ch=getchar();
 9     while(ch>='0'&&ch<='9'){num=(num<<3)+(num<<1)+ch-'0';ch=getchar();}
10     return num;
11 }
12 int main(){
13     int  q;
14     scanf("%d",&q);
15     while(q--){
16         n=get_num();
17         for(int i=1;i<=n;i++){
18             a[i]=get_num();b[i]=get_num();
19         }
20         
21         
22         f[1][0]=0;
23         f[1][1]=b[1];f[1][2]=2*b[1];
24 
25         for(int i=2;i<=n;i++){
26             f[i][0]=f[i][1]=f[i][2]=1e18;
27             for(long long k=0;k<3;k++)
28                 for(long long  j=0;j<3;j++){
29                     if(a[i]+k!=a[i-1]+j){
30                         f[i][k]=min(f[i-1][j]+k*b[i],f[i][k]);
31                     }
32                 }
33         }
34         long long ans=min(f[n][0],f[n][1]);
35         if(f[n][2]<ans)ans=f[n][2];
36         printf("%lld
",ans);
37     }
38 }

[codeforces1221D] Make The Fence Great Again dp

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