bzoj 2806 多个串匹配

题意：

多个主串和多个询问串，每次询问将询问串分成多个连续子串，如果一个子串长度>=L且在主串中出现过就是熟悉的

如果熟悉的字符串长度>=询问串长的90%就是熟悉的文章；求成为熟悉的文章的最大的L

 

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主串建广义SAM然后二分L判断可行性

使用DP判断L是否可行，一定要注意是长度不是数量

len[i]表示i位置之前最大公共长度，和spoj LCS一样...

f[i]表示前i个字符最长熟悉长度，f[i]=max{f[i-1],f[j]+j-i|i-len[i]<=j<=i-L}，显然i-len[i]不严格递增，使用单调队列优化到O(n)

//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=1100010;const int maxx=2e5+100;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;int root,last;int tots;  //总结点int l;  //字符串长度int sv[maxn*2];//int dp[maxn*2];char s[maxn];//LL sum[maxn*2];struct sam_node{    int fa,son[3];    int len;    void init(int _len)    {        len = _len;        fa = -1;        memset(son,-1,sizeof(son));    }}t[maxn*2];void sam_init(){    tots = 0;    root = last = ++tots;    t[tots].init(0);}void extend(int w){    //int w=ch-'a';    int p=last;    int np=++tots;t[tots].init(t[p].len+1);    sv[l]=np;    int q,nq;    while(p!=-1&&t[p].son[w]==-1){t[p].son[w]=np;p=t[p].fa;}    if (p==-1) t[np].fa=root;    else    {        q=t[p].son[w];        if (t[p].len+1==t[q].len){t[np].fa=q;}        else        {            nq=++tots;t[nq].init(0);            t[nq]=t[q];            t[nq].len=t[p].len+1;            t[q].fa=nq;t[np].fa=nq;            while(p!=-1&&t[p].son[w]==q){t[p].son[w]=nq;p=t[p].fa;}        }    }    last=np;}int w[maxn], r[maxn*2];//w 一倍就够void topsort(){    for(int i = 0; i <= l; ++i) w[i] = 0;    for(int i = 1; i <= tots; ++i) w[t[i].len]++;    for(int i = 1; i <= l; ++i) w[i] += w[i-1];    for(int i = tots; i >= 1; --i) r[w[t[i].len]--] = i;    r[0] = 0;}int n,m;int len[maxn],f[maxn];void getLen(){    int u=root,sum=0;    FOR(1,n,i)    {        int c=s[i]-'0';        if(t[u].son[c]!=-1) u=t[u].son[c],sum++;        else        {            W(u&&t[u].son[c]==-1) u=t[u].fa;            //cout<<u<<endl;            if(!u) u=root,sum=0;            else sum=t[u].len+1,u=t[u].son[c];        }        len[i]=sum;        //printf("len %d %d\n",i,len[i]);    }}int q[maxn],head,tail;bool check(int L){    //printf("L %d\n",L);    head=1;tail=0;    for(register int i=1;i<=n;i++)    {        f[i]=f[i-1];        if(i-L<0) continue;        while(head<=tail&&f[q[tail]]-q[tail]<f[i-L]-i+L) tail--;        q[++tail]=i-L;        while(head<=tail&&q[head]<i-len[i]) head++;        if(head<=tail) f[i]=max(f[i],f[q[head]]+i-q[head]);        //printf("f %d %d\n",i,f[i]);    }    //printf("f %d %d\n",n,f[n]);    return f[n]*10>=n*9;}void solve(){    getLen();    int L=0,R=n,ans=0;    while(L<=R)    {        int mid=(L&R)+((L^R)>>1);        if(check(mid)) ans=mid,L=mid+1;        else R=mid-1;    }    print(ans);}int main(){    //freopen( "in.txt" , "r" , stdin );    s_2(m,n);    sam_init();    FOR(1,n,i)    {        scanf("%s",s);        l=strlen(s);        last=root;        FOr(0,l,j)            extend(s[j]-'0');    }    FOR(1,m,i)    {        scanf("%s",s+1);        n=strlen(s+1);        solve();    }}