CF704B. Ant Man

Solution

经典 $d p$ ，第一次见好像是在 $Z J O I$ 的某题？
先按 $x$ 排序
用 $f [i] [j]$ 表示放入前 $i$ 个数，有 $j$ 个端点（不算边界点）的最小代价。
每次可以：
1.合并两段折线
2.制造一段新的折线
3.延长一个线段
需要对于向上和向下分别考虑贡献，特判端点情况即可。

时间复杂度 $O(n^2)$ 。

Code

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int MAXN=5005;
const int INF=0x3f3f3f3f;
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
ll x[MAXN],a[MAXN],b[MAXN],c[MAXN],d[MAXN],f[MAXN][MAXN];
signed main()
{int n=read(),S=read(),T=read();for (int i=1;i<=n;i++) x[i]=read();for (int i=1;i<=n;i++) a[i]=read();for (int i=1;i<=n;i++) b[i]=read();for (int i=1;i<=n;i++) c[i]=read();for (int i=1;i<=n;i++) d[i]=read();for (int i=0;i<=n;i++)for (int j=0;j<=n;j++) f[i][j]=loo; f[0][0]=0;for (int i=1;i<=n;i++)for (int j=(i!=1);j<n;j++){if (i==S){if (j<n) upmin(f[i][j+1],f[i-1][j]-x[i]+d[i]);if (j>0) upmin(f[i][j-1],f[i-1][j]+x[i]+c[i]);}else if (i==T){if (j<n) upmin(f[i][j+1],f[i-1][j]-x[i]+b[i]);if (j>0) upmin(f[i][j-1],f[i-1][j]+x[i]+a[i]);}else{if (j<n-1) upmin(f[i][j+2],f[i-1][j]-x[i]*2+b[i]+d[i]);if (j>1) upmin(f[i][j-2],f[i-1][j]+x[i]*2+a[i]+c[i]);if (j>1||i>T) upmin(f[i][j],f[i-1][j]+b[i]+c[i]);if (j>1||i>S) upmin(f[i][j],f[i-1][j]+a[i]+d[i]); }}printf("%lld\n",f[n][0]);return 0;
}