CF1120C Compress String

Love_xyh

于 2020-09-21 20:05:02 发布

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分类专栏：后缀数组 ST表递推

本文链接：https://blog.youkuaiyun.com/Dove_xyh/article/details/108717739

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后缀数组

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本文介绍了一种通过后缀数组和区间最小值查询优化最长公共后缀问题的算法实现，对比了朴素算法的时间复杂度，并给出了具体的代码实现。

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朴素算法n^2求最长公共后缀

#include <bits/stdc++.h>
using namespace std;
const int N=5e3+5;
int n,a,b;
int f[N],g[N][N];
char str[N];

int main(){
	scanf("%d%d%d",&n,&a,&b);
	scanf("%s",str+1);
	for (register int i=1; i<n; ++i)
	for (register int j=i+1; j<=n; ++j) 
	if (str[i]==str[j]) g[i][j]=g[i-1][j-1]+1;
	else g[i][j]=0;
	 
	memset(f,60,sizeof(f));
	f[1]=a;
	for (register int i=2; i<=n; ++i)
	{
		f[i]=f[i-1]+a;
		for (register int j=1; j<i; ++j)
		if (j<i-g[j][i]+1) f[i]=min(f[i],f[i-g[j][i]]+b);
	}
	
	printf("%d\n",f[n]);
return 0;	
}

现在我们可以通过SA来完成这项操作。同时用st表维护height数组，得到答案。

#include <bits/stdc++.h>
using namespace std;
const int N=5e3+5;
int n,m,a,b;
int sum[N],rk[N],rk2[N],tp[N],sa[N],height[N];
char s[N];
int LOG[N],minn[N][14],f[N];

inline void qsort()
{
	for (register int i=0; i<=m; ++i) sum[i]=0;
	for (register int i=1; i<=n; ++i) sum[rk[i]]++;
	for (register int i=1; i<=m; ++i) sum[i]+=sum[i-1];
	for (register int i=n; i>=1; --i) sa[sum[rk[tp[i]]]]=tp[i],sum[rk[tp[i]]]--;
}

inline void SA()
{
	m=130;
	for (register int i=1; i<=n; ++i) rk[i]=s[i]-'0'+1,tp[i]=i;
	qsort();
	int p=0;
	for (register int len=1; p<n; m=p,len<<=1)
	{
		p=0;
		for (register int i=n-len+1; i<=n; ++i) tp[++p]=i;
		for (register int i=1; i<=n; ++i) if (sa[i]>len) tp[++p]=sa[i]-len;
		qsort();
		memcpy(rk2,rk,sizeof(rk2));
		p=1; rk[sa[1]]=1;
		for (register int i=2; i<=n; ++i)
		{
			if (rk2[sa[i]]==rk2[sa[i-1]] && rk2[sa[i]+len]==rk2[sa[i-1]+len]) p=p; else p++;
			rk[sa[i]]=p;
		}
	}
}

inline void LCP()
{
	for (register int i=1; i<=n; ++i) rk[sa[i]]=i;
	int k=0;
	for (register int i=1; i<=n; ++i)
	{
		if (rk[i]==1) continue;
		if (k) k--;
		int j=sa[rk[i]-1];
		while (i+k<=n && j+k<=n && s[i+k]==s[j+k]) k++;
		height[rk[i]]=k; 
	}
//	for (register int i=1; i<=n; ++i) printf("%d ",height[i]); puts("");
}

inline void ST()
{
	LOG[0]=-1;
	for (register int i=1; i<=n; ++i) LOG[i]=LOG[i>>1]+1;
	for (register int i=1; i<=n; ++i) minn[i][0]=height[i];
	for (register int j=1; j<=13; ++j) 
	for (register int i=1; i+(1<<j)-1<=n; ++i)
	{
		minn[i][j]=min(minn[i][j-1],minn[i+(1<<(j-1))][j-1]);	
	}
}

inline void dp()
{
	memset(f,60,sizeof(f));
	f[0]=0;
	for (register int i=0; i<n; ++i)
	{
		f[i+1]=min(f[i+1],f[i]+a);
		int maxn=0;
		for (register int j=1; j<=i; ++j)
		{
			int x=rk[j],y=rk[i+1];
			if (x>y) swap(x,y);
			int s=LOG[y-(x+1)+1];
			maxn=max(maxn,min(min(minn[x+1][s],minn[y-(1<<s)+1][s]),i-j+1));
		}
		f[i+maxn]=min(f[i+maxn],f[i]+b);
	}
}

int main(){
	scanf("%d%d%d",&n,&a,&b);
	scanf("%s",s+1);
	SA();
	LCP();
	ST();
	dp();
	printf("%d\n",f[n]);
return 0;	
}