luogu2885

P2885 [USACO07NOV]电话线Telephone Wire

给出若干棵树的高度，你可以进行一种操作：把某棵树增高h，花费为h*h。

操作完成后连线，两棵树间花费为高度差*定值c。

求两种花费加和最小值。

输入输出样例

输入 #1复制

5 2
2
3
5
1
4

输出 #1复制

15

sol:显然暴力的dp很容易得到，dp[i][j]表示第i个高度为j个最小代价

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll S=; char ch=' '; bool f=;
    while(!isdigit(ch)) {f|=(ch=='-'); ch=getchar();}
    while(isdigit(ch)) {S=(S<<)+(S<<)+(ch-''); ch=getchar();}
    return (f)?(-S):(S);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<) {putchar('-'); x=-x;}
    if(x<) {putchar(x+''); return;}
    write(x/); putchar(x%+'');
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=,M=,inf=0x3f3f3f3f;
int n,c,hig[N],mx,ans;
int dp[N][M];
inline int sqr(int x){return x*x;}
int main()
{
    freopen("luogu2885_data.in","r",stdin);
    int i,j,k;
    R(n); R(c);
    for(i=;i<=n;i++)
    {
        R(hig[i]); mx=max(mx,hig[i]);
    }
    memset(dp,,sizeof dp);
    dp[][hig[]]=;
    for(i=;i<=mx-hig[];i++) dp[][hig[]+i]=i*i;
    for(i=;i<=n;i++)
    {
        for(j=hig[i];j<=mx;j++) for(k=hig[i-];k<=mx;k++)
        {
            dp[i][j]=min(dp[i][j],dp[i-][k]+sqr(j-hig[i])+c*abs(j-k));
        }
    }
    ans=inf;
    for(i=hig[n];i<=mx;i++) ans=min(ans,dp[n][i]);
    Wl(ans);
    return ;
}

然后发现容易拆开abs分类讨论一下即可

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=; bool f=; char ch=' ';
    while(!isdigit(ch))    {f|=(ch=='-'); ch=getchar();}
    while(isdigit(ch)) {s=(s<<)+(s<<)+(ch^); ch=getchar();}
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<) {putchar('-'); x=-x;}
    if(x<) {putchar(x+''); return;}
    write(x/); putchar((x%)+'');
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=,inf=0x3f3f3f3f;
int n,c,mx,hig[N];
int dp[N][],mn[];
inline int sqr(int x){return x*x;}
int main()
{
    freopen("luogu2885_data.in","r",stdin);
    int i,j;
    R(n); R(c);
    for(i=;i<=n;i++) mx=max(mx,hig[i]=read());
    memset(dp,,sizeof dp);
    for(i=hig[];i<=mx;i++) dp[][i]=sqr(i-hig[]);
    for(i=;i<=n;i++)
    {
        for(j=;j<=hig[i-]-;j++) mn[j]=inf;
        for(j=hig[i-];j<=mx;j++) mn[j]=min(mn[j-],dp[i-][j]-c*j);
        for(j=hig[i];j<=mx;j++) dp[i][j]=min(dp[i][j],mn[j]+sqr(j-hig[i])+c*j);
        mn[mx+]=inf;
        for(j=mx;j>=hig[i-];j--) mn[j]=min(mn[j+],dp[i-][j]+c*j);
        for(j=hig[i-]-;j>=;j--) mn[j]=mn[hig[i-]];
        for(j=hig[i];j<=mx;j++) dp[i][j]=min(dp[i][j],mn[j]+sqr(j-hig[i])-c*j);
//        for(j=hig[i];j<=mx;j++) cout<<i<<' '<<j<<' '<<dp[i][j]<<endl; putchar('\n');
    }
    int ans=inf;
    for(j=hig[n];j<=mx;j++) ans=min(ans,dp[n][j]);
    Wl(ans);
    return ;
}

ac