传送门
文章目录
- 题意：
- 思路：
题意：
思路：
首先证明一个结论：一个数最多被加两次。
首先假设a[i]=a[i−1]a[i]=a[i-1]a[i]=a[i−1]或a[i]=a[i+1]a[i]=a[i+1]a[i]=a[i+1]，那么此时可以将a[i]=a[i]+1a[i]=a[i]+1a[i]=a[i]+1，若此时a[i]=a[i−1]a[i]=a[i-1]a[i]=a[i−1]或a[i]=a[i+1]a[i]=a[i+1]a[i]=a[i+1]，那么可以将其再加一，这个时候一定与两边的都不想等，此时再加就没有意义了。
既然只能加两次，那么定义f[i][j]f[i][j]f[i][j]表示到了第iii个位置，这个数加了jjj次后满足条件的最小代价，转移方程：f[i][j]=min(f[i][j],f[i−1][k]+j∗b[i])f[i][j]=min(f[i][j],f[i-1][k]+j*b[i])f[i][j]=min(f[i][j],f[i−1][k]+j∗b[i])
注意初始化的时候要将其设为尽可能大的数。
代码中101010改成333即可，当时是猜了个结论。
// Problem: D. Make The Fence Great Again
// Contest: Codeforces - Educational Codeforces Round 73 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1221/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=300010,mod=1e9+7,INF=0x3f3f3f3f;
const LL inf=0x3f3f3f3f3f3f3f3f;
const double eps=1e-6;int n;
int a[N],b[N];
LL f[N][20];int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);int _; scanf("%d",&_);while(_--) {scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d%d",&a[i],&b[i]);for(int i=1;i<=n;i++) for(int j=0;j<=10;j++) f[i][j]=inf;for(int i=0;i<=10;i++) f[1][i]=1ll*b[1]*i;for(int i=2;i<=n;i++) {for(int j=0;j<=10;j++) {for(int k=0;k<=10;k++) {if(a[i-1]+k!=a[i]+j) f[i][j]=min(f[i][j],f[i-1][k]+1ll*j*b[i]);}}}LL ans=inf;for(int i=0;i<=10;i++) ans=min(ans,f[n][i]);printf("%lld\n",ans);}return 0;
}
/**/
