先考虑80分做法，即满足A串长度均不小于B串，容易发现每个B串对应的所有A串在后缀数组上都是一段连续区间，线段树优化连边然后判环求最长链即可。场上就写了这个。
100分也没有什么本质区别，没有A串长度不小于B串的性质后，区间连边变成了矩形连边，用主席树或KDTree优化连边即可，当然主席树会更靠谱，这里写了KDTree，在loj上T掉了。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 200010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int T,n,na,nb,m;
char s[N];
int rk[N<<1],tmp[N<<1],sa[N],sa2[N],Cnt[N],h[N],f[N][20],LG2[N];
struct data{int l,r,i;
}a[N],b[N];
void make()
{
int m=26;
memset(Cnt,0,sizeof(Cnt));
memset(rk,0,sizeof(rk));
memset(tmp,0,sizeof(tmp));
for (int i=1;i<=n;i++) Cnt[rk[i]=(s[i]-'a'+1)]++;
for (int i=1;i<=m;i++) Cnt[i]+=Cnt[i-1];
for (int i=n;i>=1;i--) sa[Cnt[rk[i]]--]=i;
for (int k=1;k<=n;k<<=1)
{
int p=0;
for (int i=n-k+1;i<=n;i++) sa2[++p]=i;
for (int i=1;i<=n;i++) if (sa[i]>k) sa2[++p]=sa[i]-k;
for (int i=1;i<=m;i++) Cnt[i]=0;
for (int i=1;i<=n;i++) Cnt[rk[i]]++;
for (int i=1;i<=m;i++) Cnt[i]+=Cnt[i-1];
for (int i=n;i>=1;i--) sa[Cnt[rk[sa2[i]]]--]=sa2[i];
for (int i=1;i<=n*2;i++) tmp[i]=rk[i];
p=1,rk[sa[1]]=1;
for (int i=2;i<=n;i++)
{
if (tmp[sa[i]]!=tmp[sa[i-1]]||tmp[sa[i]+k]!=tmp[sa[i-1]+k]) p++;
rk[sa[i]]=p;
}
m=p;if (m==n) break;
}
for (int i=1;i<=n;i++)
{
h[i]=max(h[i-1]-1,0);
while (s[i+h[i]]==s[sa[rk[i]-1]+h[i]]) h[i]++;
}
for (int i=1;i<=n;i++) f[i][0]=h[sa[i]];
for (int j=1;j<20;j++)
for (int i=1;i<=n;i++)
f[i][j]=min(f[i][j-1],f[min(n,i+(1<<j-1))][j-1]);
}
int query(int x,int y)
{
x++;if (x>y) return N;
return min(f[x][LG2[y-x+1]],f[y-(1<<LG2[y-x+1])+1][LG2[y-x+1]]);
}
int root,cnt,p[N<<1],degree[N<<1],q[N<<1],val[N<<1],t;
ll F[N<<1];
struct data2{int to,nxt;
}edge[N<<9];
struct data3{int l,r,x,y,ch[2],i;
}tree[N<<1];
void addedge(int x,int y){t++;degree[y]++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
ll topsort()
{
int head=0,tail=0;
for (int i=1;i<=cnt;i++)
{
F[i]=val[i];
if (!degree[i]) q[++tail]=i;
}
while (head<tail)
{
int x=q[++head];
for (int i=p[x];i;i=edge[i].nxt)
{
degree[edge[i].to]--;
F[edge[i].to]=max(F[edge[i].to],F[x]+val[edge[i].to]);
if (!degree[edge[i].to]) q[++tail]=edge[i].to;
}
}
if (tail<cnt) return -1;
ll ans=0;
for (int i=1;i<=cnt;i++) ans=max(ans,F[i]);
return ans;
}
int newnode(){cnt++;p[cnt]=degree[cnt]=val[cnt]=F[cnt]=0;return cnt;}
bool cmp1(const data&a,const data&b)
{
return a.l<b.l||a.l==b.l&&a.r<b.r;
}
bool cmp2(const data&a,const data&b)
{
return a.r<b.r||a.r==b.r&&a.l<b.l;
}
bool cmp3(const data&a,const data&b)
{
return a.i<b.i;
}
void build(int &k,int l,int r)
{
if (l>r) return;
k=newnode();
int mid=l+r>>1;
bool flag;
/*if (rand()&1)
{
long double s1=0,s2=0,u=0,v=0;
for (int i=l;i<=r;i++) s1+=1ll*a[i].l*a[i].l,s2+=a[i].l;
u=s1-s2/(r-l+1)*s2;s1=0,s2=0;
for (int i=l;i<=r;i++) s1+=1ll*a[i].r*a[i].r,s2+=a[i].r;
v=s1-s2/(r-l+1)*s2;
flag=u>v;
}
else*/
{
int u=a[l].l,v=a[l].l,w;
for (int i=l;i<=r;i++) u=min(u,a[i].l),v=max(v,a[i].l);
w=v-u;
u=a[l].r,v=a[l].r;for (int i=l;i<=r;i++) u=min(u,a[i].r),v=max(v,a[i].r);
flag=v-u<w;
}
if (flag) nth_element(a+l,a+mid,a+r+1,cmp1);
else nth_element(a+l,a+mid,a+r+1,cmp2);
tree[k].i=a[mid].i;
tree[k].l=tree[k].r=a[mid].l;
tree[k].x=tree[k].y=a[mid].r;
tree[k].ch[0]=tree[k].ch[1]=0;
for (int i=l;i<=r;i++)
{
tree[k].l=min(tree[k].l,a[i].l);
tree[k].r=max(tree[k].r,a[i].l);
tree[k].x=min(tree[k].x,a[i].r);
tree[k].y=max(tree[k].y,a[i].r);
}
build(tree[k].ch[0],l,mid-1);
build(tree[k].ch[1],mid+1,r);
if (tree[k].ch[0]) addedge(k,tree[k].ch[0]);
if (tree[k].ch[1]) addedge(k,tree[k].ch[1]);
addedge(k,tree[k].i);
}
void cnnct(int k,int l,int r,int x,int y,int p)
{
if (!k||l>tree[k].r||r<tree[k].l||x>tree[k].y||y<tree[k].x) return;
if (l<=tree[k].l&&r>=tree[k].r&&x<=tree[k].x&&y>=tree[k].y) {addedge(p,k);return;}
if (l<=a[tree[k].i].l&&r>=a[tree[k].i].l&&x<=a[tree[k].i].r&&y>=a[tree[k].i].r) addedge(p,tree[k].i);
cnnct(tree[k].ch[0],l,r,x,y,p);
cnnct(tree[k].ch[1],l,r,x,y,p);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
const char LL[]="%I64d
";
#else
const char LL[]="%lld
";
#endif
T=read();srand(time(0));
for (int i=2;i<=200000;i++)
{
LG2[i]=LG2[i-1];
if ((2<<LG2[i])<=i) LG2[i]++;
}
while (T--)
{
scanf("%s",s+1);
n=strlen(s+1);
make();
na=read();
for (int i=1;i<=na;i++)
{
a[i].l=read(),a[i].r=read(),a[i].i=i;
a[i].r=a[i].r-a[i].l+1,a[i].l=rk[a[i].l];
}
nb=read();
for (int i=1;i<=nb;i++)
{
b[i].l=read(),b[i].r=read(),b[i].i=b[i].r-b[i].l+1;
int x=rk[b[i].l];b[i].l=b[i].r=x;
int l=1,r=x;
while (l<=r)
{
int mid=l+r>>1;
if (query(mid,x)>=b[i].i) b[i].l=mid,r=mid-1;
else l=mid+1;
}
l=x,r=n;
while (l<=r)
{
int mid=l+r>>1;
if (query(x,mid)>=b[i].i) b[i].r=mid,l=mid+1;
else r=mid-1;
}
}
for (int i=1;i<=na;i++) F[i]=degree[i]=p[i]=0,val[i]=a[i].r;t=0;cnt=na;
build(root,1,na);sort(a+1,a+na+1,cmp3);
m=read();
for (int i=1;i<=m;i++)
{
int x=read(),y=read();
cnnct(root,b[y].l,b[y].r,b[y].i,n,x);
}
cout<<topsort()<<endl;
}
return 0;
}
更好的做法是直接对给出的A串按字典序排序，显然也可以通过SA做到，这样同样用线段树优化连边即可。因为没有什么本质区别就不写了。