题解【洛谷P2264】情书

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;
}