[codeforces]F. Compress String

　　题解:  SAM板子题 但是要做一个小dp

　　很显然  有个dp就是

　　$ dp[i]=min(dp[i],dp[j]+b) $若在子串[1,j-1]中子串[j,i]存在 

　　那么我们就用SAM维护每个子串最早出现在字符串中的位置  然后check一下就行了

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define link(x) for(edge *j=h[x];j;j=j->next)
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,r,l) for(int i=r;i>=l;i--)
const int MAXN=3e5+10;
const double eps=1e-8;
#define ll long long
using namespace std;
const int inf=1e9;
struct edge{int t;edge*next;}e[MAXN<<1],*h[MAXN],*o=e;
void add(int x,int y){o->t=y;o->next=h[x];h[x]=o++;}
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return x*f;
}


int n,a,b;
char str[MAXN];
int fa[MAXN],dis[MAXN],ch[MAXN][26],minn[MAXN];
int cnt,cur,rt;
void built(int x){
    int last=cur;cur=++cnt;dis[cur]=dis[last]+1;int p=last;minn[cur]=dis[cur];
    for(;p&&!ch[p][x];p=fa[p])ch[p][x]=cur;
    if(!p)fa[cur]=rt;
    else {
	int q=ch[p][x];
	if(dis[q]==dis[p]+1)fa[cur]=q;
	else{
	    int nt=++cnt;dis[nt]=dis[p]+1;minn[nt]=inf;
	    memcpy(ch[nt],ch[q],sizeof(ch[q]));
	    fa[nt]=fa[q];fa[q]=fa[cur]=nt;
	    for(;ch[p][x]==q;p=fa[p])ch[p][x]=nt;
	}
    }
}

void dfs(int x){
    link(x){dfs(j->t);minn[x]=min(minn[x],minn[j->t]);}
}

int dp[MAXN];

int main(){
    n=read();a=read();b=read();
    scanf("%s",str+1);
    cnt=cur=rt=1;minn[rt]=inf;
    inc(i,1,n)built(str[i]-'a');
    inc(i,1,cnt)add(fa[i],i);
    dfs(rt);
    inc(i,1,n)dp[i]=inf;
    inc(i,1,n){
	int temp=rt;
	inc(j,i,n){
	    if(j==i)dp[j]=min(dp[j],dp[j-1]+a);
	    int t=str[j]-'a';temp=ch[temp][t];
	    if(minn[temp]<i)dp[j]=min(dp[j],dp[i-1]+b);
	}
    }
    printf("%d\n",dp[n]);
}