HOJ 12823 Swyper Keyboard （模拟）

给你一个特殊的键盘，打字是通过滑行来打两个不同的字母的。现在给你这些打出的字母，找出打字时经过的其他字母。然后一些nc要求。

暴搞，或者自己枚举所有可能性。。。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <assert.h>
#include <algorithm>
#define MAX 1234567890
#define MIN -1234567890
#define eps 1e-8

using namespace std;

int cnt;
char key[104];
char road[20004];

/// * A B C D E * 00 01 02 03 04 05 06 ** 00 01 02 03 04 **
/// F G H I J K L 07 08 09 10 11 12 13 05 06 07 08 09 10 11
/// M N O P Q R S 14 15 16 17 18 19 20 12 13 14 15 16 17 18
/// T U V W X Y Z 21 22 23 24 25 26 27 19 20 21 22 23 24 25

int P[28][2] = {
{2, 8}, {4, 8}, {6, 8}, {8, 8}, {10, 8}, {12, 8}, {14, 8},
{2, 6}, {4, 6}, {6, 6}, {8, 6}, {10, 6}, {12, 6}, {14, 6},
{2, 4}, {4, 4}, {6, 4}, {8, 4}, {10, 4}, {12, 4}, {14, 4},
{2, 2}, {4, 2}, {6, 2}, {8, 2}, {10, 2}, {12, 2}, {14, 2},
};

int pos[26] = {
1, 2, 3, 4, 5,
7, 8, 9, 10, 11, 12, 13,
14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25, 26, 27,
};

int ret[28] = {
0, 0, 1, 2, 3, 4, 0,
5, 6, 7, 8, 9, 10, 11,
12, 13, 14, 15, 16, 17, 18,
19, 20, 21, 22, 23, 24, 25,
};

int judge(int x1, int y1, int x2, int y2)
{
return x1*y2 - x2*y1;
}

///给定起点位置和终点位置
void findroad(int first, int last)
{
int dx = P[last][0] > P[first][0] ? 1 : -1;
int dy = P[last][1] > P[first][1] ? 1 : -1;
if(P[last][0] == P[first][0])
{
if(dy > 0) for(int i = first-7; i >= last; i -= 7) road[cnt++] = 'A' + ret[i];
if(dy < 0) for(int i = first+7; i <= last; i += 7) road[cnt++] = 'A' + ret[i];
}
else if(P[last][1] == P[first][1])
{
if(dx > 0) for(int i = first+1; i <= last; i += 1) road[cnt++] = 'A' + ret[i];
if(dx < 0) for(int i = first-1; i >= last; i -= 1) road[cnt++] = 'A' + ret[i];
}
else
{
int next = first;
int sx = P[last][0] - P[first][0];
int sy = P[last][1] - P[first][1];
while((P[next][0] != P[last][0] || P[next][1] != P[last][1]) && next >= 0 && next <= 27)
{
int tmpx = P[next][0] + dx;
int tmpy = P[next][1] + dy;
int tmpj = judge(sx, sy, tmpx - P[first][0], tmpy - P[first][1]);
if(dx < 0 && dy < 0)
{
if(tmpj < 0) next += 7;
if(tmpj > 0) next -= 1;
if(tmpj == 0) next += 6;
}
if(dx > 0 && dy < 0)
{
if(tmpj < 0) next += 1;
if(tmpj > 0) next += 7;
if(tmpj == 0) next += 8;
}
if(dx > 0 && dy > 0)
{
if(tmpj < 0) next -= 7;
if(tmpj > 0) next += 1;
if(tmpj == 0) next -= 6;
}
if(dx < 0 && dy > 0)
{
if(tmpj < 0) next -= 1;
if(tmpj > 0) next -= 7;
if(tmpj == 0) next -= 8;
}
if(next != 0 && next != 6) road[cnt++] = 'A' + ret[next];
}
}
}

int main()
{
#ifdef BellWind
freopen("12823.in", "r", stdin);
#endif // BellWind

int T;
scanf("%d", &T);

while(T--)
{
int n;
scanf("%d %s", &n, key);
cnt = 0;
memset(road, 0, sizeof(road));
int len = strlen(key);
road[cnt++] = key[0];
for(int l = 0; l+1 < len; l++) findroad(pos[key[l]-'A'], pos[key[l+1]-'A']);
road[cnt] = '\0';
// puts(road);
bool flag = true;
for(int t = 0; t < n; t++)
{
char word[104];
scanf("%s", word);
if(flag)
{
int lenw = strlen(word);
int k, l;
for(l = 0, k = 0; l < cnt && k < lenw; l++) if(word[k] == road[l]) k++;
if(k == lenw)
{
flag = false;
printf("%s\n", word);
}
}
}
if(flag) printf("NO SOLUTION\n");
}

return 0;
}