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f_{t,i,j} = MAX(f_{t-1,x',y'})+1

复杂度:O(Tnm),只能过50%

看到k \le 200那么我们是不是可以直接用?

f_{k,i,j}为在区间k中在位置(i,j)能最多滑行多远

f_{k,i,j} = MAX(f_{k-1,i,j},f_{k-1,i',j'}+dis_{(i,j),(i',j'))})

那么复杂度就是O(kn^3),可以用单调队列优化掉其中一个n

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;i++)
    #define Fdr(i,x,y) for(int i=x;i>=y;i--)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) (int(log2(n)))
    #define inf 0x3f3f3f3f
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 211
int n,m,X,Y,K,u,len,ans=0,f[N][N],q[N],p[N];
int dx[5]={0,-1,1,0,0},dy[5]={0,0,0,-1,1};
char a[N][N];
void work(int x,int y){
    int h=1,t=0;
    for(int i=1;x>=1&&x<=n&&y>=1&&y<=m;i++,x+=dx[u],y+=dy[u])
    if(a[x][y]=='x')h=1,t=0;
    else{
        while(h<=t&&q[t]+i-p[t]<f[x][y])t--;
        p[++t]=i;
        q[t]=f[x][y];
        if(p[t]-p[h]>len)h++;

        f[x][y]=q[h]+i-p[h];

        ans=MAX(ans,f[x][y]);
    }
}
int main(){
    in>>n>>m>>X>>Y>>K;
    Fur(i,1,n)in>>(a[i]+1);
    int s,t;
    clr(f,128);
    f[X][Y]=0;
    Fur(k,1,K){
        in>>s>>t>>u;
        len=t-s+1;
        if(u==1)Fur(i,1,m)work(n,i);
        if(u==2)Fur(i,1,m)work(1,i);
        if(u==3)Fur(i,1,n)work(i,m);
        if(u==4)Fur(i,1,n)work(i,1);
    }
    out<<ans<<ln;
}
LG 2254 [NOI2005]瑰丽华尔兹
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