题面

看到每一单词在同一句话中出现多次感动值不叠加，一眼想到 \(\text{set}\)。

首先将词汇列表中的单词存储起来，我用的是 \(\text{set}\)。

对于每一个句子的单词，我们可以先判断它是否是词汇列表中的单词，然后将它加入一个 \(\text{set}\) 中，如果扫描到句号就将答案加上 \(\text{set}\) 的大小，然后清空 \(\text{set}\) 。

#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d\n", __FUNCTION__, __LINE__)
#define itn int
#define gI gi

using namespace std;

typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;

inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return f * x;
}

inline LL gl()
{
    LL f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return f * x;
}

int n, ans;
string xsl, pp;
set <string> td, kk; 

int main()
{
    //freopen("1.in", "r", stdin);
    //freopen(".out", "w", stdout);
    n = gi();
    for (int i = 1; i <= n; i+=1)
    {
        cin >> xsl;
        int len = xsl.size();
        for (int j = 0; j < len; j+=1)
        {
            if (xsl[j] >= 'A' && xsl[j] <= 'Z') xsl[j] = xsl[j] - 'A' + 'a';
        }
        td.insert(xsl);
    }
    char c = getchar();
    xsl = "";
    while (~scanf("%c", &c))
    {
        //puts("!");
        if (c == '\n') break;
        if (c >= 'A' && c <= 'Z') c = c - 'A' + 'a';
        if (c >= 'a' && c <= 'z') xsl = xsl + c;
        else 
        {
            if (xsl != "") 
            {
                if (td.find(xsl) != td.end()) kk.insert(xsl);
                xsl = "";
            }
            if (c == '.') 
            {
                ans += kk.size(), kk.clear();
            }
        }
    }
    printf("%d\n", ans);
    return 0;
}