$ NOI2005 $ magnificent waltz

\ (NOI2005 \) magnificent Waltz

Monotonous queue achieve good question.

Set \ (f [k] [i ] [j] \) represents the \ (K \) segment interval \ ((i, j) \ ) an optimal answer.

\ (F [K] [I] [J] = max (F [K-. 1] [the I] [J] + Dis) \) ( \ (Dis \) is the distance, \ ((the I, J) \) is a legitimate position. too lazy to write ...)

Consider adding queue monotonous rolling array solve (MLE, TLE \) \ problems. . .

Clear the queue can be in the face of obstacles. . .

Note that the \ (f [i] [j ] \) before the addition of the queue, the answer can not be updated, because our \ (max \) is the \ (k-1 \) a \ (max \) , skillet after the update. . .

#include<bits/stdc++.h>
using namespace std;
inline int read()
{
    int f=1,w=0;char x=0;
    while(x<'0'||x>'9') {if(x=='-') f=-1; x=getchar();}
    while(x!=EOF&&x>='0'&&x<='9') {w=(w<<3)+(w<<1)+(x^48);x=getchar();}
    return w*f;
}
const int N=250;
struct Group{int Val,Stp;} Q[N];
int n,m,Sx,Sy,K,ans,Mp[N][N],f[N][N];
int dx[5]={0,-1,1,0,0},dy[5]={0,0,0,-1,1};
inline void Dp(int X,int Y,int S,int D)
{
    int H=1,T=0;
    for(int Stp=1;1<=X&&X<=n&&1<=Y&&Y<=m;Stp++,X+=dx[D],Y+=dy[D])
        if(!Mp[X][Y]) H=1,T=0;
        else
        {
            while(H<=T&&Q[T].Val+Stp-Q[T].Stp<f[X][Y]) T--;
            Q[++T]=(Group){f[X][Y],Stp};
            while(H<=T&&Q[T].Stp-Q[H].Stp>S) H++;
            f[X][Y]=Q[H].Val+Stp-Q[H].Stp;ans=max(ans,f[X][Y]);
        }
}
int main(){
#ifndef ONLINE_JUDGE
    freopen("A.in","r",stdin);
#endif
    n=read(),m=read(),Sx=read(),Sy=read(),K=read();
    for(int i=1;i<=n;i++)
    {
        string S;cin>>S;
        for(int j=0;j<m;j++)
            if(S[j]=='.') Mp[i][j+1]=1;
            else Mp[i][j+1]=0;
    }
    memset(f,0xf3,sizeof(f));f[Sx][Sy]=0;
    for(int i=1;i<=K;i++)
    {
        int L=read(),R=read(),D=read();
        if(D==1) for(int j=1;j<=m;j++) Dp(n,j,R-L+1,D);
        if(D==2) for(int j=1;j<=m;j++) Dp(1,j,R-L+1,D);
        if(D==3) for(int j=1;j<=n;j++) Dp(j,m,R-L+1,D);
        if(D==4) for(int j=1;j<=n;j++) Dp(j,1,R-L+1,D);
    }
    printf("%d\n",ans);
}