Codeforces Round #344 (Div. 2) D. Messenger （KMP）
D. Messenger
time limit per test2 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard output
Each employee of the "Blake Techologies" company uses a special messaging app "Blake Messenger". All the stuff likes this app and uses it constantly. However, some important futures are missing. For example, many users want to be able to search through the message history. It was already announced that the new feature will appear in the nearest update, when developers faced some troubles that only you may help them to solve.
All the messages are represented as a strings consisting of only lowercase English letters. In order to reduce the network load strings are represented in the special compressed form. Compression algorithm works as follows: string is represented as a concatenation of n blocks, each block containing only equal characters. One block may be described as a pair (li, ci), where li is the length of the i-th block and ci is the corresponding letter. Thus, the string s may be written as the sequence of pairs .
Your task is to write the program, that given two compressed string t and s finds all occurrences of s in t. Developers know that there may be many such occurrences, so they only ask you to find the number of them. Note that p is the starting position of some occurrence of s in t if and only if tptp + 1...tp + |s| - 1 = s, where ti is the i-th character of string t.
Note that the way to represent the string in compressed form may not be unique. For example string "aaaa" may be given as , , ...
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 200 000) — the number of blocks in the strings t and s, respectively.
The second line contains the descriptions of n parts of string t in the format "li-ci" (1 ≤ li ≤ 1 000 000) — the length of the i-th part and the corresponding lowercase English letter.
The second line contains the descriptions of m parts of string s in the format "li-ci" (1 ≤ li ≤ 1 000 000) — the length of the i-th part and the corresponding lowercase English letter.
Output
Print a single integer — the number of occurrences of s in t.
Examples
inputCopy
5 3
3-a 2-b 4-c 3-a 2-c
2-a 2-b 1-c
outputCopy
1
inputCopy
6 1
3-a 6-b 7-a 4-c 8-e 2-a
3-a
outputCopy
6
inputCopy
5 5
1-h 1-e 1-l 1-l 1-o
1-w 1-o 1-r 1-l 1-d
outputCopy
0
Note
In the first sample, t = "aaabbccccaaacc", and string s = "aabbc". The only occurrence of string s in string t starts at position p = 2.
In the second sample, t = "aaabbbbbbaaaaaaacccceeeeeeeeaa", and s = "aaa". The occurrences of s in t start at positions p = 1, p = 10, p = 11, p = 12, p = 13 and p = 14.
题意：
就是找一个字符串在另一个字符串出现的次数。
思路：
我们在读入的时候把相邻的并且相等的字符合并在一起，然后当m<=2我们直接暴力来做，
当 m>=3使，我们去掉s字符串的第一个块和最后一个块，中间的部分建立next 双指标的数组，然后与t字符串进行双指标的kmp算法，当匹配到的时候，我们看匹配的前一个位置和s字符串的第一个块是否匹配，以及后一个位置和s字符串的最后一个块是否匹配，即是否字符相等，如果字符相等，个数是否大于等于。
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 200010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
pair<ll,char> a[maxn];
pair<ll,char> b[maxn];
pair<ll,char> d[maxn];
int n,m;
ll ans=0ll;
int nextarr[maxn];
void build()
{
nextarr[0]=-1;
int i=0;
int j=-1;
while(i<m)
{
if(j==-1||(b[i].se==b[j].se&&b[i].fi==b[j].fi))
{
j++;
i++;
nextarr[i]=j;
}else
{
j=nextarr[j];
}
}
}
std::vector<pii> v;
void kmp()
{
int x=0,y=0;
while(x<n&&y<m)
{
if(a[x]==b[y])
{
x++;
y++;
}
else if(y==0)
{
x++;
}else
{
y=nextarr[y];
}
if(y==m)
{
v.push_back(mp(x-m,x+1));
y=nextarr[y];
}
}
}
int main()
{
//freopen("D:\\code\\text\\input.txt","r",stdin);
//freopen("D:\\code\\text\\output.txt","w",stdout);
scanf("%d %d",&n,&m);
ll temp;
char c;
repd(i,1,n)
{
scanf("%lld-%c",&temp,&c);
if(a[i-1].se==c)
{
a[i-1].fi+=temp;
i--;
n--;
}else
{
a[i].fi=temp;
a[i].se=c;
}
}
repd(i,1,m)
{
scanf("%lld-%c",&temp,&c);
if(b[i-1].se==c)
{
b[i-1].fi+=temp;
i--;
m--;
}else
{
b[i].fi=temp;
b[i].se=c;
}
}
if(m==1)
{
repd(i,1,n)
{
if(a[i].se==b[1].se&&a[i].fi>=b[1].fi)
{
ans+=a[i].fi-b[1].fi+1ll;
}
}
cout<<ans<<endl;
}else if(m==2)
{
repd(i,1,n)
{
if(a[i].se==b[1].se&&a[i].fi>=b[1].fi&&a[i+1].se==b[2].se&&a[1+i].fi>=b[2].fi)
{
ans++;
}
}
cout<<ans<<endl;
}else
{
repd(i,1,m)
{
d[i]=b[i];
}
repd(i,1,m-2)
{
b[i]=b[i+1];
}
m-=2;
repd(i,0,m-1)
{
b[i]=b[i+1];
}
repd(i,0,n-1)
{
a[i]=a[i+1];
}
build();
kmp();
for(auto x:v)
{
// cout<<(a[x.fi-1].se==d[x.fi].se)<<" "<<(a[x.se-1].se==d[x.se].se)<<" "<<(a[x.fi-1].fi>=d[x.fi].fi)<<" "<<(a[x.se-1].fi>=d[x.se].fi)<<endl;
if(a[x.fi-1].se==d[1].se&&a[x.se-1].se==d[m+2].se&&a[x.fi-1].fi>=d[1].fi&&a[x.se-1].fi>=d[m+2].fi)
{
ans++;
}
// cout<<x.fi<<" "<<x.se<<endl;
}
cout<<ans<<endl;
}
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}
