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CodeForces 713C Sonya and Problem Wihtout a Legend

$dp$。

非严格递增的可以由$dp$解决。可以将严格递增转化为非严格递增。

要保证$a[i]<a[i+1]$,就是要保证$a[i]<=a[i+1]-1$,也就是$a[i]-i≤a[i+1]-1-i$,等价于$a[i]-i≤a[i+1]-(i+1)$。

因此只要将$a[i]-i$,求非严格递增的就可以了。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<iostream>using namespace std;typedef long long LL;const double pi=acos(-1.0),eps=1e-6;void File(){    freopen("D:\\in.txt","r",stdin);    freopen("D:\\out.txt","w",stdout);}template <class T>inline void read(T &x){    char c=getchar(); x=0;    while(!isdigit(c)) c=getchar();    while(isdigit(c)) {x=x*10+c-0; c=getchar();}}const int maxn=3000+10;long long dp[maxn][maxn];long long a[maxn],q[maxn];long long MIN[maxn];int n;long long ABS(long long a){    return max(a,-a);}bool cmp(const int&a,const int&b){    return a>b;}int main(){    while(~scanf("%d",&n))    {        for(int i=1; i<=n; i++)        {            scanf("%lld",&q[i]);            a[i]=q[i];            q[i]=q[i]-i;            a[i]=a[i]-i;        }        sort(q+1,q+n+1);        memset(dp,-1,sizeof dp);        memset(MIN,0,sizeof MIN);        for(int i=1; i<=n; i++)        {            for(int j=1; j<=n; j++) dp[i][j]=MIN[j]+ABS(a[i]-q[j]);            MIN[1]=dp[i][1];            for(int j=2; j<=n; j++) MIN[j]=min(MIN[j-1],dp[i][j]);        }        long long ans1=dp[n][1];        for(int i=2; i<=n; i++) ans1=min(ans1,dp[n][i]);        printf("%lld\n",ans1);    }    return 0;}

 

CodeForces 713C Sonya and Problem Wihtout a Legend