NOIP 模拟 $86; m 矩阵$

题解 (by;zjvarphi)

现特判，如果存在相邻的点相等，那么它可以来回走，所以直接输出 -1。

那么现在直接暴力搜索，可以发现最多只会走 (log) 步。

但是这样状态数还是太多，记忆化一下即可。

Code

#include<bits/stdc++.h>
#define ri signed
#define pd(i) ++i
#define bq(i) --i
#define func(x) std::function<x>
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    #define debug1(x) std::cerr << #x"=" << x << ' '
    #define debug2(x) std::cerr << #x"=" << x << std::endl
    #define Debug(x) assert(x)
    struct nanfeng_stream{
        template<typename T>inline nanfeng_stream &operator>>(T &x) {
            bool f=false;x=0;char ch=gc();
            while(!isdigit(ch)) f|=ch=='-',ch=gc();
            while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc();
            return x=f?-x:x,*this;
        }
    }cin;
}
using IO::cin;
namespace nanfeng{
    #define ch(x,y) (x-1)*m+(y)
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    static const int N=4e4+7;
    struct edge{int v,nxt;}e[N<<2];
    int first[N],w[N],dp[N][330],t=1,n,m,al,ans;
    int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
    bool fg;
    auto add=[](int u,int v) {e[t]={v,first[u]},first[u]=t++;};
    func(int(int,int)) dfs=[](int x,int nw) {
        if (nw<=300&&dp[x][nw]) return dp[x][nw];
        int cur=0;
        for (ri i(first[x]),v;i;i=e[i].nxt)
            if (1ll*nw*w[x]==1ll*w[v=e[i].v]) cur=cmax(cur,dfs(v,nw));
            else if (!nw&&!(w[e[i].v]%w[x])) cur=cmax(cur,dfs(v,w[e[i].v]/w[x]));
        if (nw<=300) dp[x][nw]=cur+1;
        return cur+1;
    };
    inline int main() {
        FI=freopen("matrix.in","r",stdin);
        FO=freopen("matrix.out","w",stdout);
        cin >> n >> m;
        for (ri i(1);i<=n;pd(i))
            for (ri j(1);j<=m;pd(j)) {
                int cur=ch(i,j);
                cin >> w[cur];
            }
        for (ri i(1);i<=n;pd(i))
            for (ri j(1);j<=m;pd(j)) {
                int cur=ch(i,j);
                for (ri k(0);k<4;pd(k)) {
                    int ti=i+dx[k],tj=j+dy[k],nw=ch(ti,tj);
                    if (ti<1||ti>n||tj<1||tj>m) continue;
                    add(cur,nw);
                }
            }
        al=n*m;
        for (ri i(1);i<=al;pd(i))
            for (ri j(first[i]);j;j=e[j].nxt) if (w[i]==w[e[j].v]) {fg=true;break;}
        if (fg) return printf("-1"),0;
        for (ri i(1);i<=al;pd(i)) ans=cmax(ans,dfs(i,0));
        printf("%d
",ans);
        return 0;
    }
}
int main() {return nanfeng::main();}