给定一张包含N个单词的表，每个单词有个价值W。要求从中选出一个子序列使得其
中的每个单词是后一个单词的子串，最大化子序列中W的和。

第一行一个整数TEST，表示数据组数。 
接下来TEST组数据，每组数据第一行为一个整数N。 
接下来N行，每行为一个字符串和一个整数W。

TEST行，每行一个整数，表示W的和的最大值。 
 
 
数据规模 
设字符串的总长度为Len 
30.的数据满足，TEST≤5，N≤500，Len≤10^4 
100.的数据满足，TEST≤10，N≤20000，Len≤3*10^5

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 using namespace std;
  6 #define maxn 300020
  7
  8 typedef long long ll;
  9 struct node{
 10     int ls,rs;
 11     ll mx,tag;
 12 }sgt[maxn * 2];
 13 struct node2{
 14     int next,to;
 15 }e[maxn * 2];
 16 int head[maxn],cnt;
 17 char ch[maxn];
 18 int T,n;
 19 int nxt[maxn][26],tot,dfstime,fa[maxn];
 20 int fail[maxn],rt,cur,l[maxn],r[maxn],rec[maxn],w[maxn];
 21 int q[maxn],tt,hh;
 22 ll f[maxn],ans;
 23
 24 inline void clear(){
 25     for (int i = 1 ; i <= n ; i++) f[i] = 0;
 26     for (int i = 0 ; i <= tot ; i++) fail[i] = l[i] = r[i] = head[i] = fa[i] = rec[i] = 0;
 27     for (int i = 1 ; i <= tot ; i++) e[i].next = e[i].to = 0;
 28     for (int i = 0 ; i <= tot ; i++) memset(nxt[i],0,sizeof(nxt[i]));
 29     for (int i = 1 ; i <= cnt ; i++) sgt[i].tag = sgt[i].mx = 0 , sgt[i].ls = sgt[i].rs = 0;
 30     tot = cnt = dfstime = 0;
 31     ans = 0;
 32 }
 33 inline void adde(int x,int y){
 34     e[++cnt].to = y;
 35     e[cnt].next = head[x];
 36     head[x] = cnt;
 37 }
 38 inline void insert(int x){
 39     if ( !nxt[cur][x] ) nxt[cur][x] = ++tot , fa[tot] = cur;
 40     cur = nxt[cur][x];
 41 }
 42 void dfs(int x){
 43     if ( x ) l[x] = ++dfstime;
 44     for (int i = head[x] ; i ; i = e[i].next) dfs(e[i].to);
 45     r[x] = dfstime;
 46 }
 47 void build(int &x,int l,int r){
 48     x = ++cnt;
 49     if ( l == r ) return;
 50     int mid = (l + r) >> 1;
 51     build(sgt[x].ls,l,mid);
 52     build(sgt[x].rs,mid + 1,r);
 53 }
 54 void build_AC(){
 55     tt = hh = 0;
 56     for (int i = 0 ; i < 26 ; i++) if ( nxt[0][i] ) q[tt++] = nxt[0][i];
 57     while ( hh < tt ){
 58         int x = q[hh++];
 59         for (int i = 0 ; i < 26 ; i++){
 60             if ( nxt[x][i] ){
 61                 int p = fail[x];
 62                 while ( !nxt[p][i] && p ) p = fail[p];
 63                 fail[nxt[x][i]] = nxt[p][i]; //不是p
 64                 q[tt++] = nxt[x][i];
 65             }
 66         }
 67     }
 68     for (int i = 1 ; i <= tot ; i++) adde(fail[i],i);
 69     dfs(0);
 70     cnt = 0;
 71     build(rt,1,tot);
 72 }
 73 inline void update(int x){
 74     sgt[x].mx = max(sgt[sgt[x].ls].mx,sgt[sgt[x].rs].mx);
 75 }
 76 inline void add(int x,ll d){
 77     sgt[x].tag = max(sgt[x].tag,d);
 78     sgt[x].mx = max(sgt[x].mx,d);
 79 }
 80 inline void pushdown(int x){
 81     if ( sgt[x].tag != 0 ){
 82         add(sgt[x].ls,sgt[x].tag);
 83         add(sgt[x].rs,sgt[x].tag);
 84         sgt[x].tag = 0;
 85     }
 86 }
 87 void modify(int x,int l,int r,int ls,int rs,ll d){
 88     if ( ls <= l && rs >= r ){
 89         add(x,d);
 90         return;
 91     }
 92     pushdown(x);
 93     int mid = (l + r) >> 1;
 94     if ( ls <= mid ) modify(sgt[x].ls,l,mid,ls,rs,d);
 95     if ( rs > mid ) modify(sgt[x].rs,mid + 1,r,ls,rs,d);
 96     update(x);
 97 }
 98 ll query(int x,int l,int r,int id){
 99     if ( l == r ) return sgt[x].mx;
100     pushdown(x);
101     int mid = (l + r) >> 1;
102     if ( id <= mid ) return query(sgt[x].ls,l,mid,id);
103     return query(sgt[x].rs,mid + 1,r,id);
104 }
105 void Dodp(){
106     for (int i = 1 ; i <= n ; i++){
107         int p = rec[i];
108         while ( p ){
109             f[i] = max(f[i],query(rt,1,tot,l[p]));
110             p = fa[p];
111         }
112         f[i] += w[i];
113         f[i] = max(0ll,f[i]);
114         modify(rt,1,tot,l[rec[i]],r[rec[i]],f[i]);
115     }
116     for (int i = 1 ; i <= n ; i++) ans = max(ans,f[i]);
117 }
118 int main(){
119     freopen("input.txt","r",stdin);
120     scanf("%d",&T);
121     while ( T-- ){
122         clear();
123         scanf("%d",&n);
124         for (int i = 1 ; i <= n ; i++){
125             scanf("%s",ch) , scanf("%d",&w[i]);
126             int l = strlen(ch);
127             cur = 0;
128             for (int j = 0 ; j < l ; j++){
129                 insert(ch[j] - 'a');
130             }
131             rec[i] = cur;
132         }
133         build_AC();
134         Dodp();
135         printf("%lld\n",ans);
136     }
137     return 0;
138 }