【BZOJ1499】瑰丽华尔兹（动态规划）

题面

BZOJ

题解

先写部分分
设\(f[t][i][j]\)表示当前在\(t\)时刻，位置在\(i,j\)时走的最多的步数
这样子每一步要么停要么走
时间复杂度\(O(nmt)\)
得分\(40~70\)分
（据说这样能过？？？）

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
#define RG register
#define MAX 210
inline int read()
{RG int x=0,t=1;RG char ch=getchar();while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();if(ch=='-')t=-1,ch=getchar();while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();return x*t;
}
char g[MAX][MAX];
int ans,n,m,X,Y,K,L[MAX],R[MAX],D[MAX];
int d[5][2]={0,0,-1,0,1,0,0,-1,0,1};
int f[2][MAX][MAX],T[MAX*MAX];
int main()
{n=read();m=read();X=read();Y=read();K=read();for(int i=1;i<=n;++i)scanf("%s",g[i]+1);for(int i=1;i<=K;++i)L[i]=read(),R[i]=read(),D[i]=read();for(int i=1;i<=K;++i)for(int j=L[i];j<=R[i];++j)T[j]=D[i];memset(f,-1,sizeof(f));f[0][X][Y]=0;int nw=1,pw=0;for(int tt=1;tt<=R[K];++tt,nw^=1,pw^=1){for(int i=1;i<=n;++i)for(int j=1;j<=m;++j)f[nw][i][j]=-1;for(int i=1;i<=n;++i)for(int j=1;j<=m;++j){if(f[pw][i][j]==-1)continue;int xx=i+d[T[tt]][0],yy=j+d[T[tt]][1];f[nw][i][j]=max(f[nw][i][j],f[pw][i][j]);if(xx<1||yy<1||xx>n||yy>m)continue;if(g[xx][yy]=='x')continue;f[nw][xx][yy]=max(f[nw][xx][yy],f[pw][i][j]+1);}}for(int i=1;i<=n;++i)for(int j=1;j<=m;++j)ans=max(ans,f[R[K]&1][i][j]);printf("%d\n",ans);return 0;
}

发现转移可以用单调队列优化
于是分四种情况进行讨论
用单调队列优化转移即可

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
#define RG register
#define MAX 210
inline int read()
{RG int x=0,t=1;RG char ch=getchar();while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();if(ch=='-')t=-1,ch=getchar();while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();return x*t;
}
char g[MAX][MAX];
int ans,n,m,X,Y,K,L[MAX],R[MAX],D[MAX];
int d[5][2]={0,0,-1,0,1,0,0,-1,0,1};
int f[2][MAX][MAX],T[MAX*MAX];
int Q[MAX],h,t;
int main()
{n=read();m=read();X=read();Y=read();K=read();for(int i=1;i<=n;++i)scanf("%s",g[i]+1);for(int i=1;i<=K;++i)L[i]=read(),R[i]=read(),D[i]=read();for(int i=1;i<=n;++i)for(int j=1;j<=m;++j)f[0][i][j]=-1e9;f[0][X][Y]=0;int nw=1,pw=0;for(int tt=1;tt<=K;++tt,nw^=1,pw^=1){for(int i=1;i<=n;++i)for(int j=1;j<=m;++j)f[nw][i][j]=f[pw][i][j];if(D[tt]==1)for(int j=1;j<=m;++j){h=1;t=0;for(int i=n;i;--i){if(g[i][j]=='x'){h=1;t=0;continue;}while(h<=t&&Q[h]-i>R[tt]-L[tt]+1)++h;while(h<=t&&f[pw][Q[t]][j]+Q[t]<=f[pw][i][j]+i)--t;Q[++t]=i;if(h<=t)f[nw][i][j]=f[pw][Q[h]][j]+Q[h]-i;}}else if(D[tt]==2)for(int j=1;j<=m;++j){h=1;t=0;for(int i=1;i<=n;++i){if(g[i][j]=='x'){h=1;t=0;continue;}while(h<=t&&i-Q[h]>R[tt]-L[tt]+1)++h;while(h<=t&&f[pw][Q[t]][j]-Q[t]<=f[pw][i][j]-i)--t;Q[++t]=i;if(h<=t)f[nw][i][j]=f[pw][Q[h]][j]+i-Q[h];}}else if(D[tt]==3)for(int i=1;i<=n;++i){h=1;t=0;for(int j=m;j;--j){if(g[i][j]=='x'){h=1;t=0;continue;}while(h<=t&&Q[h]-j>R[tt]-L[tt]+1)++h;while(h<=t&&f[pw][i][Q[t]]+Q[t]<=f[pw][i][j]+j)--t;Q[++t]=j;if(h<=t)f[nw][i][j]=f[pw][i][Q[h]]+Q[h]-j;}}elsefor(int i=1;i<=n;++i){h=1;t=0;for(int j=1;j<=m;++j){if(g[i][j]=='x'){h=1;t=0;continue;}while(h<=t&&j-Q[h]>R[tt]-L[tt]+1)++h;while(h<=t&&f[pw][i][Q[t]]-Q[t]<=f[pw][i][j]-j)--t;Q[++t]=j;if(h<=t)f[nw][i][j]=f[pw][i][Q[h]]+j-Q[h];}}}for(int i=1;i<=n;++i)for(int j=1;j<=m;++j)ans=max(ans,f[K&1][i][j]);printf("%d\n",ans);return 0;
}