bzoj 3790: 神奇项链

IQ--

其实就是用manacher搞出来一些回文的线段，然后就是判断如何用最少的线段覆盖整个区间就行了，

区间覆盖的话，DP一下，用BIT优化一下。

#include <bits/stdc++.h>
#define inf 1e9
using namespace std;

const int maxn=100005;

int n,m,cnt;
int p[maxn<<1],c[maxn];
char ch[maxn],a[maxn<<1];
struct seg{
    int l,r;
}l[maxn];

int lowbit(int x) {return x&(-x);}
int query(int x)
{
    if (!x) return 0;
    int tmp=inf;
    for (; x<=n; x+=lowbit(x)) tmp=min(c[x],tmp);
    return tmp;
}
void change(int x, int val)
{
    for (;x;x-=lowbit(x)) c[x]=min(c[x],val);
}

bool operator < (seg a, seg b)
{
    return a.r<b.r;
}
void add(int x, int y)
{
    x=x/2+1; y=y/2-1;
    if (x>y) return;
    l[++cnt]=(seg){x,y};
}
void manachar()
{
    m=2*n+1;
    for (int i=1; i<=n; i++) a[i<<1]=ch[i],a[i<<1|1]='#';
    a[0]='+'; a[m+1]='-'; a[1]='#';
    int mx=0,id=0;
    for (int i=1; i<=m; i++)
    {
        if (mx>i) p[i]=min(mx-i,p[2*id-i]); else p[i]=1;
        while (a[i+p[i]]==a[i-p[i]]) p[i]++;
        add(i-p[i],i+p[i]);
        if (p[i]+i>mx) mx=p[i]+i,id=i;
    }
}

int main()
{
    while (scanf("%s",ch+1)!=EOF)
    {
        cnt=0;
        n=strlen(ch+1); for (int i=1; i<=n; i++) c[i]=inf;
        manachar(); sort(l+1,l+cnt+1);
        int ans=inf; 
    //    for (int i=1; i<=cnt; i++) printf("%d %d
",l[i].l,l[i].r);
        for (int i=1; i<=cnt; i++)
        {
            int x=query(l[i].l-1)+1;
            change(l[i].r,x);
            if (l[i].r==n) ans=min(ans,x);
        }
        cout<<ans-1<<endl;
    }
    return 0;
}