分析

BFS出到每一个点的（距离Limit）最短路径
将收到的宝藏sort+unique
然后再暴力max一下，即可

代码

//Newcoder 18 C
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
#include <cstring>

#define LL long long
#define Lowbit(X) (X&(-X))
#define Lson (X<<1)
#define Rson (X<<1|1)
#define Cl(X,Y) memset((X),(Y),sizeof(X))
#define FOR(i,A,B) for(LL i=A;i<=B;i++)
#define BOR(i,A,B) for(LL i=A;i>=B;i--)
#define FOR_SIDE(i,A) for(LL i=Head[A];i;i=Next[i])
#define INF 0x3f3f3f3f3f3f3f3f
using namespace std;
const LL MAXN=1e3+10;

LL Row,Line,Ans,Test;
LL Cube[MAXN][MAXN],Stax,Stay,Dist,Limit;
bool Vis[MAXN][MAXN],Jud[MAXN][MAXN];
LL Num[1000010],Cnt,Dep[MAXN][MAXN];
LL Dx[5]={0,0,-1,0,1},Dy[5]={0,1,0,-1,0};
struct Node { LL E,F; };
queue<Node>Temp;

inline void File() {
    freopen(".in","r",stdin);
    freopen(".out","w",stdout);
}

inline void BFS() {
    Temp.push(Node{ Stax,Stay });
    Vis[Stax][Stay]=true;
    while(!Temp.empty()) {
        Node New=Temp.front();
        Temp.pop();
        if(Cube[New.E][New.F]>0) Num[++Cnt]=Cube[New.E][New.F];
        if(Dep[New.E][New.F]>=Dist) continue;
        FOR(i,1,4) {
            LL Nx=New.E+Dx[i],Ny=New.F+Dy[i];
            if(Nx>Row || Nx<1 || Ny>Line || Ny<1 || Vis[Nx][Ny]) continue;
            Dep[Nx][Ny]=Dep[New.E][New.F]+1;
            Vis[Nx][Ny]=true;
            Temp.push(Node{ Nx,Ny });
        }
    }
}

inline void Solve() {
    sort(Num+1,Num+Cnt+1);
    LL Size=unique(Num+1,Num+Cnt+1)-(Num+1);
    if(Size<Limit) { printf("no\n"); }
    else {
        FOR(i,1,Size-Limit+1) 
            Ans=min(Ans,Num[i+Limit-1]-Num[i]);
        printf("%lld\n",Ans);
    }
}

int main() {
    //File();
    scanf("%lld",&Test);
    while(Test--) {
        Cl(Num,0); Cnt=0;
        Cl(Vis,false); Cl(Cube,0);
        Ans=INF; Cl(Dep,0); Cl(Jud,false);
        while(!Temp.empty()) Temp.pop();
        scanf("%lld %lld %lld %lld %lld %lld",&Row,&Line,&Stax,&Stay,&Dist,&Limit);
        FOR(i,1,Row) FOR(j,1,Line) {
            scanf("%lld",&Cube[i][j]);
            if(Cube[i][j]==-1) Vis[i][j]=true;
        }
        if(Cube[Stax][Stay]==-1) { cout<<"no"<<endl; continue; }
        BFS();
        Solve();
    }
    //fclose(stdin); fclose(stdout);
   // system("pause");
    return 0;
}