Bzoj 4567: [Scoi2016] Back Word dictionary tree + greedy

Test Instructions

Meaning is to put the word, each word must be placed before the suffix of the word first put up, the cost of each word is equal to the position of the word minus the last occurrence of the word suffix position
N<=100000,sigma (len) <=510000 Analysis

The first idea is to use a suffix array to build the map, if you want to really fight it is estimated that T. And also did not think how greedy ... is still too weak.

In fact, as long as a dictionary tree can handle the problem of building maps.
And then talk ... Every time I find a smaller subtree to go greedy. Prove it. It's not obvious. Code

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <

Algorithm> #define N 100005 using namespace std;
int cnt,n,tim,size[n*10],last[n*20],ch[n*10][30],top,sta[n],val[n*10],tot,dfn[n*10];
Long long ans; struct Edge{int to,next;}
E[n];

Char s[n*10];
    void ins () {int Len=strlen (s), now=1;
        for (int i=len-1;i>=0;i--) if (ch[now][s[i]-' a ']) now=ch[now][s[i]-' a '];
    Else now=ch[now][s[i]-' a ']=++tot;
Val[now]=1;

} void Addedge (int u,int v) {e[++cnt].to=v;e[cnt].next=last[u];last[u]=cnt;}
    void build (int x,int y) {if (val[x]) Addedge (y,x), y=x;
for (int i=0;i<26;i++) if (Ch[x][i]) build (ch[x][i],y);
    } void Get_size (int x) {size[x]=1;
for (int i=last[x];i;i=e[i].next) get_size (e[i].to), size[x]+=size[e[i].to];

} BOOL cmp (int a,int b) {return size[a]<size[b];}
    void Dfs (int x) {dfn[x]=++tim;int l=top+1;
 for (int i=last[x];i;i=e[i].next) sta[++top]=e[i].to;   if (top<l) return;
    Sort (sta+l,sta+top+1,cmp);
    for (int i=l;i<=top;i++) DFS (Sta[i]), ans+=dfn[sta[i]]-dfn[x];
TOP=L-1;
    } int main () {scanf ("%d", &n);
    Tot=1;
        for (int i=1;i<=n;i++) {scanf ("%s", s);
    Ins ();
    } build (n);
    Get_size (1);
    DFS (1);
    printf ("%lld", ans);
return 0; }