Luogu P2662 Cattle Farm Fence

Transmission Door

Since a plank can be cut off up to \ (m\), it is possible to pre-preprocess which lengths of planks are available, open a bucket, then mark the \ ([l-m,l]\) , and then remove the marked number

Assuming the available length \ (a_1,a_2,,, a_n\) from small to large order, we can first without \ (a_1\), only with the length of the back to spell, and then with \ (a_1\) together, so set \ (di_i\) for the shortest length (\ (di_0=0\)) that can be mod\ with the back and \ (a_1\ ) for \ (i\) , then the \ (a_2,,, a_n\) can be taken from the location i\ \ ( (i+a_j) mod\ a_1\), Edge ( a_j\) Edge, run a shortest way to find all \ (di\)

The final answer is \ (\max_{i=0}^{a_1-1}di_i-a_1\), because these lengths \ (d_i\) do not use \ (a_1\) to spell out, this inside any one length minus \ (a_1\) is the length that cannot be spelled out

Pay attention to the sentence without solution

#include <bits/stdc++.h> #define LL long long#define il inline#define re registerusing namespace Std;const int n=3000    +10,m=n*n;il LL Rd () {re LL x=0,w=1;re char ch; while (ch< ' 0 ' | |    Ch> ' 9 ') {if (ch== '-') W=-1;ch=getchar ();}    while (ch>= ' 0 ' &&ch<= ' 9 ') {x= (x<<3) + (x<<1) + (ch^48); Ch=getchar ();} return x*w;} int to[m],nt[m],w[m],hd[n],tot=1;il void Add (int x,int y,int z) {++tot,to[tot]=y,nt[tot]=hd[x],w[tot]=z,hd[x]=tot;}  struct node{int x,d; BOOL operator < (const node &b) Const {return d>b.d;}}; int n,m,a[n],di[n];short Vv[n];il int gcd (int a,int b) {return B?GCD (b,a%b): A;}  int main () {n=rd (), M=rd ();      while (n--) {int x=rd ();    ++vv[max (1,x-m)],--vv[x+1];  } n=0;      for (int i=1;i<=n-10;i++) {vv[i]+=vv[i-1];    if (vv[i]>0) a[++n]=i;  } int gg=a[1];  for (int i=2;i<=n;i++) GG=GCD (Gg,a[i]); if (gg>1| |  A[1]==1) {puts ("-1"); return 0;} No solution for (int i=0;i<a[1];i++) for (int j=2;j<=n;j++) Add (i, (i+A[J])%a[1],a[j]);  memset (di,63,sizeof (DI));  di[0]=0;  Priority_queue<node> Q;  Q.push (node) {0,0});      while (!q.empty ()) {int x=q.top (). X,d=q.top (). D;      Q.pop ();      if (d>di[x]) continue;          for (int i=hd[x];i;i=nt[i]) {int y=to[i];              if (Di[y]>di[x]+w[i]) {di[y]=di[x]+w[i];            Q.push (node) {y,di[y]});  }}} int ans=a[1];  for (int i=0;i<a[1];i++) Ans=max (Ans,di[i]);  printf ("%d\n", ans-a[1]); return 0;}

Luogu P2662 Cattle field fence