Codeforces Round #299 (Div. 2) D. Tavas and Malekas kmp

题目链接：

http://codeforces.com/problemset/problem/535/D

D. Tavas and Malekas

time limit per test2 seconds
memory limit per test256 megabytes #### 问题描述 > Tavas is a strange creature. Usually "zzz" comes out of people's mouth while sleeping, but string s of length n comes out from Tavas' mouth instead. > > > Today Tavas fell asleep in Malekas' place. While he was sleeping, Malekas did a little process on s. Malekas has a favorite string p. He determined all positions x1 < x2 < ... < xk where p matches s. More formally, for each xi (1 ≤ i ≤ k) he condition sxisxi + 1... sxi + |p| - 1 = p is fullfilled. > > Then Malekas wrote down one of subsequences of x1, x2, ... xk (possibly, he didn't write anything) on a piece of paper. Here a sequence b is a subsequence of sequence a if and only if we can turn a into b by removing some of its elements (maybe no one of them or all). > > After Tavas woke up, Malekas told him everything. He couldn't remember string s, but he knew that both p and s only contains lowercase English letters and also he had the subsequence he had written on that piece of paper. > > Tavas wonders, what is the number of possible values of s? He asked SaDDas, but he wasn't smart enough to solve this. So, Tavas asked you to calculate this number for him. > > Answer can be very large, so Tavas wants you to print the answer modulo 109 + 7. #### 输入 > The first line contains two integers n and m, the length of s and the length of the subsequence Malekas wrote down (1 ≤ n ≤ 106 and 0 ≤ m ≤ n - |p| + 1). > > The second line contains string p (1 ≤ |p| ≤ n). > > The next line contains m space separated integers y1, y2, ..., ym, Malekas' subsequence (1 ≤ y1 < y2 < ... < ym ≤ n - |p| + 1). #### 输出 > In a single line print the answer modulo 1000 000 007. ####样例输入 > 6 2 > ioi > 1 3

样例输出

26

题意

给你一个子串p，且在长度为n的原串中出现的m个插入位，问原串有几种可能。

题解

我们只要考虑插入位置之间的重叠部分是否能够完全匹配，这个其实就是p串的前缀在和后缀匹配，用kmp处理出failed指针就可以轻松解决了。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1e6+10;

const int mod=1e9+7;

LL xp26[maxn];

char str[maxn];
int arr[maxn];
int n,m,len;

int f[maxn],vis[maxn];
void getFail(char* P,int* f){
    int m=strlen(P);
    f[0]=0; f[1]=0;
    for(int i=1;i<m;i++){
        int j=f[i];
        while(j&&P[i]!=P[j]) j=f[j];
        f[i+1]=P[i]==P[j]?j+1:0;
    }

    clr(vis,0);
    int j=f[m];
    while(j){
        vis[j]=1;
//        bug(j);
        j=f[j];
    }
}

int main() {

    scf("%d%d",&n,&m);
    scf("%s",str);
    len=strlen(str);

    getFail(str,f);

    rep(i,0,m) scf("%d",&arr[i]);

    int ans=m?len:0;

    for(int i=1; i<m; i++) {
        int x1=arr[i-1],x2=arr[i];
        int y1=x1+len-1,y2=x2+len-1;
        if(y1<x2) ans+=len;
        else {
            int l=y1-x2+1;
//            bug(l);
            if(vis[l]) {
                ans+=y2-y1;
            } else {
                prf("0
");
                return 0;
            }
        }
    }

    LL last=1;
    for(int i=0;i<n-ans;i++) last*=26,last%=mod;
    prf("%I64d
",last);

    return 0;
}

//end-----------------------------------------------------------------------

写了个哈希的，死且仅死在第67组数据，xrz，

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1e6+10;

const int P=321;
const int mod=1e9+7;

unsigned long long H[maxn],xp[maxn];
LL xp26[maxn];

char str[maxn];
int arr[maxn];
int n,m,len;

void pre() {
    xp[0]=1;
    rep(i,1,maxn) xp[i]=xp[i-1]*P;

    xp26[0]=1;
    rep(i,1,maxn) xp26[i]=xp26[i-1]*26%mod;
}

void get_H() {
    H[0]=0;
    for(int i=1; i<=len; i++) {
        H[i]=H[i-1]*P+str[i]-'a'+1;
    }
}

unsigned long long query(int l,int r) {
    return H[r]-H[l-1]*xp[r-l+1];
}

int main() {
    pre();
    scf("%d%d",&n,&m);
    scf("%s",str+1);
    len=strlen(str+1);

    get_H();

    rep(i,0,m) scf("%d",&arr[i]);

    sort(arr,arr+m);

    int ans=m?len:0;

    for(int i=1; i<m; i++) {
        int x1=arr[i-1],x2=arr[i];
        int y1=x1+len-1,y2=x2+len-1;
        if(y1<x2) ans+=len;
        else {
            int l=y1-x2+1;
            if(query(1,l)==query(len-l+1,len)) {
                ans+=y2-y1;
            } else {
//                puts("fuck");
//                bug(l);
                prf("0
");
                return 0;
            }
        }
    }
//    bug(ans);
    prf("%I64d
",xp26[n-ans]);

    return 0;
}

//end-----------------------------------------------------------------------