Codeforces Round #344 (Div. 2) D. Messenger（kmp，细节）

题意：给出两个字符串所给的形式为 每个字符串的每段的字符和它的长度 求子串在主串中出现的次数

解题思路：去掉子串的头和尾后kmp计算中间部分在主串中出现的次数最后判断即可

在next跳转的时候要注意最后一位和模式串跳转位num相同（如果是模式串跳转位是1则不需要考虑）

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <list>
#include <queue>
#include <map>
#include <bitset>
using namespace std;
#define L(i) i<<1
#define R(i) i<<1|1
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-3
#define maxn 100100
#define MOD 1000000007

int n,m;
int Next[maxn<<1];
long long ans;
struct node
{
long long num;
char c;
bool operator<(const node &a)const
{
return num < a.num;
}
}st1[maxn<<1],st2[maxn<<1];

void get_next()
{
int i = 0,j = -1;
Next[0] = -1;
while(i < m)
{
if(j == -1 || st2[i].c == st2[j].c)
{
i++;
j++;
Next[i] = j;
}
else
j = Next[j];
}
}
int ok(int i,int j)
{
if(j == 0 || j == m-1)
return st1[i].num >= st2[j].num;
return st1[i].num == st2[j].num;
}
void kmp()
{
int i = 0,j = 0;
while(i < n)
{
if(j == -1 || (st1[i].c == st2[j].c && ok(i,j)))
{
i++;
j++;
}
else
j = Next[j];
if(j == m)
{

j = Next[j];
//printf("%d %d %lld %lld\n",i,j,st1[i-1].num,st2[j-1].num);
while(st1[i-1].num != st2[j-1].num && j != 1 && j)
j = Next[j];
ans++;
}
}

}
int main()
{
while(scanf("%d%d",&n,&m) != EOF)
{
int k1 = 0,k2 = 0;
for(int i = 0; i < n; i++)
{
scanf("%lld-%c",&st1[k1].num,&st1[k1].c);
if(k1 && st1[k1].c == st1[k1-1].c)
st1[k1-1].num += st1[k1].num;
else
k1++;
}
for(int i = 0; i < m; i++)
{
scanf("%lld-%c",&st2[k2].num,&st2[k2].c);
if(k2 && st2[k2].c == st2[k2-1].c)
st2[k2-1].num += st2[k2].num;
else
k2++;
}
n = k1;
m = k2;
if(m == 1)
{
for(int i = 0; i < n; i++)
if(st1[i].c == st2[0].c && st1[i].num >= st2[0].num)
ans += st1[i].num-st2[0].num+1ll;
printf("%lld\n",ans);
continue;
}
get_next();
kmp();
printf("%lld\n",ans);
}
return 0;
}