hdu 4117 GRE Words AC自动机DP
题目:给出n个串,问最多能够选出多少个串,使得前面串是后面串的子串(按照输入顺序)
分析:
其实这题是这题SPOJ 7758. Growing Strings AC自动机DP的进阶版本,主题思想差不多。
对于这题来说,需要离线操作。dp转移也是很显然。
但是由于数据比较大,所以普通的沿着fail指针往上走,逐步更新答案会TLE。
考虑把fail指针反向,由于ac自动机的每个节点均有唯一的fail指针,若是沿着fail指针往上走,显然都会走到root,所以反向之后显然是一棵树,不妨称之为fail树。fail树有什么特点呢?可以画个图试试,如果儿子节点出现过,那么他的祖先显然也会出现!
因此,正确的做法时:
建立所有单词的AC自动机,对于每个节点的转移,都是从parent[]或者从fail[],fail[fail[]],...得到的。可以看出fail[]的关系形成一棵树,于是问题转化成,不断在节点处插入,询问点到根路径上的最大值,可以利用dfs序列转化用线段树维护。
#include <queue> #include <vector> #include <cstdio> #include <cstring> #include <complex> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; typedef unsigned long long ull; #define debug puts("here") #define rep(i,n) for(int i=0;i<n;i++) #define rep1(i,n) for(int i=1;i<=n;i++) #define REP(i,a,b) for(int i=a;i<=b;i++) #define foreach(i,vec) for(unsigned i=0;i<vec.size();i++) #define pb push_back #define RD(n) scanf("%d",&n) #define RD2(x,y) scanf("%d%d",&x,&y) #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w) #define All(vec) vec.begin(),vec.end() #define MP make_pair #define PII pair<int,int> #define PQ priority_queue #define cmax(x,y) x = max(x,y) #define cmin(x,y) x = min(x,y) #define Clear(x) memset(x,0,sizeof(x)) #define lson rt<<1 #define rson rt<<1|1 #define SZ(x) x.size() /******** program ********************/ const int MAXN = 3e5+5; const int kind = 26; char s[MAXN]; int beg[MAXN]; struct segTree { int l,r,mx; int cov; inline int mid() { return (l+r)>>1; } } tree[MAXN<<2]; void build(int l,int r,int rt) { tree[rt].l = l; tree[rt].r = r; tree[rt].cov = tree[rt].mx = 0; if(l==r)return; int mid = tree[rt].mid(); build(l,mid,lson); build(mid+1,r,rson); } void update(int rt){ if(tree[rt].cov==0)return; cmax( tree[lson].cov,tree[rt].cov ); cmax( tree[rson].cov,tree[rt].cov ); cmax( tree[lson].mx,tree[rt].cov ); cmax( tree[rson].mx,tree[rt].cov ); tree[rt].cov = 0; } void modify(int l,int r,int mx,int rt) { if(l<=tree[rt].l&&tree[rt].r<=r) { cmax(tree[rt].mx,mx); cmax(tree[rt].cov,mx); return; } update(rt); int mid = tree[rt].mid(); if(r<=mid)modify(l,r,mx,lson); else if(l>mid)modify(l,r,mx,rson); else { modify(l,r,mx,lson); modify(l,r,mx,rson); } tree[rt].mx = max( tree[lson].mx,tree[rson].mx ); } int ask(int pos,int rt) { if(tree[rt].l==tree[rt].r) return tree[rt].mx; update(rt); int mid = tree[rt].mid(); if(pos<=mid)return ask(pos,lson); else return ask(pos,rson); } int ch[MAXN][kind],fail[MAXN]; int val[MAXN]; int st[MAXN],ed[MAXN]; int tot,tim; vector<int> adj[MAXN]; inline void set(int x) { Clear(ch[x]); fail[x] = 0; adj[x].clear(); } inline void init() { set(1); tot = 1; } inline int newNode() { set(++tot); return tot; } inline int ind(char c) { return c-'a'; } inline void ins(int x,int y,int val) { int r = 1; REP(i,x,y) { int c = ind(s[i]); if(ch[r][c]==0) ch[r][c] = newNode(); r = ch[r][c]; } } inline void build() { queue<int> q; q.push(1); while(!q.empty()) { int r = q.front(); q.pop(); if(fail[r]) adj[ fail[r] ].pb(r); rep(c,kind) { int x = ch[r][c]; if(!x)continue; q.push(x); int y = fail[r]; while(y&&ch[y][c]==0) y = fail[y]; fail[x] = y?ch[y][c]:1; } } } void dfs(int x) { st[x] = ++ tim; foreach(i,adj[x]) { int y = adj[x][i]; dfs(y); } ed[x] = tim; } inline int run() { init(); int n; RD(n); rep1(i,n) { scanf("%s",s+beg[i-1]); RD(val[i]); beg[i] = beg[i-1]+strlen(s+beg[i-1]); ins(beg[i-1],beg[i]-1,val[i]); } build(); tim = 0; dfs(1); build(1,tim,1); int ans = 0; rep1(i,n) { int r = 1; int now = 0; for(int j=beg[i-1]; j<beg[i]; j++) { int c = ind(s[j]); r = ch[r][c]; cmax(now,ask( st[r],1 )); } now += val[i]; modify(st[r],ed[r],now,1); cmax(ans,now); } return ans; } int main() { #ifndef ONLINE_JUDGE freopen("sum.in","r",stdin); //freopen("sum.out","w",stdout); #endif int ssize = 128 << 20; // 256MB char *ppp = (char*)malloc(ssize) + ssize; __asm__("movl %0, %%esp\n" :: "r"(ppp) ); int ncase,Ncase = 0; RD(ncase); while(ncase--) printf("Case #%d: %d\n",++Ncase,run()); return 0; }