CodeForces 631D Messenger

$KMP$。

$n=1$和$n=2$的时候可以单独计算。$n>2$时，可以拿字符和数字分别做一次匹配，然后扫描一遍判断一下就可以计算出答案了。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<bitset>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-6;
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c=getchar(); x=0;
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) {x=x*10+c-'0'; c=getchar();}
}

const int maxn=200010;
struct X { LL num; int a; }s[maxn],t[maxn];
int lens,lent,n,m;
char op[100];
int nx[maxn],f[maxn],g[maxn];
LL a[maxn],b[maxn]; int len1,len2;

void get_next()
{
    int j=-1,i=0; nx[0]=-1;
    while(i<len2)
    {
        if(j==-1||b[i]==b[j]) i++, j++, nx[i]=j;
        else j=nx[j];
    }
}

void kmp(bool x){
    get_next();
    int i=0,j=0;
    while(i<len1){
        if(j==-1||a[i]==b[j]) i++ ,j++;
        else j=nx[j];
        if(j==len2)
        {
            if(x==0) f[i-j+len2]=1;
            else g[i-j+len2]=1;
        }
    }
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%s",op); int L=strlen(op);
        int num=0,sign;
        for(int j=0;j<=L-3;j++) num=num*10+op[j]-'0';
        sign=op[L-1]-'a'+1;
        if(lens==0||sign!=s[lens-1].a) { s[lens].num=num; s[lens].a=sign; lens++; }
        else s[lens-1].num+=num;
    }

    for(int i=1;i<=m;i++)
    {
        scanf("%s",op); int L=strlen(op);
        int num=0,sign;
        for(int j=0;j<=L-3;j++) num=num*10+op[j]-'0';
        sign=op[L-1]-'a'+1;
        if(lent==0||sign!=t[lent-1].a) { t[lent].num=num; t[lent].a=sign; lent++; }
        else t[lent-1].num+=num;
    }

    if(lent==1)
    {
        LL ans=0;
        for(int i=0;i<lens;i++)
        {
            if(s[i].a!=t[0].a) continue;
            if(s[i].num<t[0].num) continue;
            ans=ans+s[i].num-t[0].num+1;
        }
        printf("%lld\n",ans);
    }
    else if(lent==2)
    {
        LL ans=0;
        for(int i=0;i<lens-1;i++)
        {
            if(s[i].a!=t[0].a||s[i+1].a!=t[1].a) continue;
            if(t[0].num>s[i].num||t[1].num>s[i+1].num) continue;
            ans++;
        }
        printf("%lld\n",ans);
    }
    else
    {
        for(int i=0;i<lens;i++) a[i]=s[i].a;
        for(int i=0;i<lent;i++) b[i]=t[i].a;
        len1=lens,len2=lent; kmp(0);

        memset(a,0,sizeof a); memset(b,0,sizeof b);
        for(int i=0;i<lens;i++) a[i]=s[i].num;
        for(int i=1;i<lent-1;i++) b[i-1]=t[i].num;
        len1=lens; len2=lent-2; kmp(1);

        LL ans=0;
        for(int i=0;i<lens;i++)
        {
            if(f[i+1]==0) continue;
            if(g[i]==0) continue;
            if(s[i].num<t[lent-1].num) continue;
            if(s[i-lent+1].num<t[0].num) continue;
            ans++;
        }
        printf("%lld\n",ans);

    }
    return 0;
}