D - Dumb feature
Gym - 102020D
代码语言:javascript复制&:字典树的模板题,根据来的串建树,再查询。不过当时没弄出来,要映射一下子,把字母映射成键盘上的数字。 ps:这题的数据应该是有问题,只统计小于100的,然后直接用 map 也可以水过,不过不建议。
//引自 lxw
#include <cstdio>
#include <algorithm>
#include <bits/stdc .h>
using namespace std;
typedef long long ll;
const int N = 1000010;
const int SIZE = 15;
int ch[N][SIZE];
int val[N],sz;
int ans;
void init() // 初始化
{
sz = 1;
memset(ch,0,sizeof(ch));
memset(val,0,sizeof(val));
}
int getId(int x)
{
if(x <= 3) return 2;
if(x <= 6) return 3;
if(x <= 9) return 4;
if(x <= 12) return 5;
if(x <= 15) return 6;
if(x <= 19) return 7;
if(x <= 22) return 8;
if(x <= 26) return 9;
}
void creat(char *s)
{
int u = 0, len = strlen(s);
for(int i = 0; i < len; i )
{
int c = getId(s[i] - 'a' 1);
if(!ch[u][c]) ch[u][c] = sz ;
u = ch[u][c];
val[u] ;
}
}
int query(char *s)
{
int u = 0, len = strlen(s);
for(int i = 0; i < len; i )
{
int c = s[i] - '0';
if(!ch[u][c]) return 0;
u = ch[u][c];
}
return val[u];
}
char s[100005];
int main()
{
int n,q;
while(~scanf("%d %d", &n, &q))
{
init();
for(int i = 0; i < n ; i )
{
getchar();
scanf("%s", s);
creat(s);
}
while(q--)
{
getchar();
scanf("%s", s);
printf("%dn",query(s));
}
}
return 0;
}
ps做法:
代码语言:javascript复制//引自 zm
#include <algorithm>
#include <cstdio>
#include <iostream>
#include <vector>
#include <cstring>
#include<map>
using namespace std;
typedef long long ll;
map<string,int> mp;
char s[100005];
char a[100005];
int main()
{
int n,m;
scanf("%d %d",&n, &m);
for(int j = 1;j <= n;j )
{
scanf("%s",a);
int len = strlen(a);
for(int i = 0;i < min(len,100);i )
{
if(a[i] == 'a'||a[i] == 'b'||a[i] == 'c') s[i] = '2';
else if(a[i] == 'd'||a[i] == 'e'||a[i] == 'f') s[i] = '3';
else if(a[i] == 'g'||a[i] == 'h'||a[i] == 'i') s[i] = '4';
else if(a[i] == 'j'||a[i] == 'k'||a[i] == 'l') s[i] = '5';
else if(a[i] == 'm'||a[i] == 'n'||a[i] == 'o') s[i] = '6';
else if(a[i] == 'p'||a[i] == 'q'||a[i] == 'r'||a[i] == 's') s[i] = '7';
else if(a[i] == 't'||a[i] == 'u'||a[i] == 'v') s[i] = '8';
else if(a[i] == 'w'||a[i] == 'x'||a[i] == 'y'||a[i] == 'z') s[i] = '9';
s[i 1] = '