URAL 1002 Phone Numbers

URAL_1002

dp的思路还是比较好想的，可以用f[i]表示递推到第i个数字的时候最短用几个单词表示。这样只要枚举当前单词的起始位置j，如果j~i可以构成一个单词，那么就尝试更新f[i]，而判断j~i是否可以构成一个单词的过程可以用哈希表实现。

#include<stdio.h>
#include<string.h>
#define INF 0x3f3f3f3f
#define HASH 1000003
#define MAXD 50010
#define MAXL 60
#define MAXN 110
int N, head[HASH], next[MAXD], f[MAXN], p[MAXN], g[MAXN], len[MAXD];
char b[MAXN], ch[300], word[MAXD][MAXL], code[MAXD][MAXL];
int hash(char *str, int a, int b)
{
int i, h = 0, seed = 131;
for(i = a; i <= b; i ++)
h = h * seed + str[i];
return (h & 0x7fffffff) % HASH;
}
void Insert(int s)
{
int h = hash(code[s], 0, len[s] - 1);
next[s] = head[h];
head[h] = s;
}
void init()
{
int i, j;
memset(head, -1, sizeof(head));
ch['o'] = ch['q'] = ch['z'] = '0';
ch['i'] = ch['j'] = '1';
ch['a'] = ch['b'] = ch['c'] = '2';
ch['d'] = ch['e'] = ch['f'] = '3';
ch['g'] = ch['h'] = '4';
ch['k'] = ch['l'] = '5';
ch['m'] = ch['n'] = '6';
ch['p'] = ch['r'] = ch['s'] = '7';
ch['t'] = ch['u'] = ch['v'] = '8';
ch['w'] = ch['x'] = ch['y'] = '9';
scanf("%d", &N);
for(i = 0; i < N; i ++)
{
scanf("%s", word[i]);
for(j = 0; word[i][j]; j ++)
code[i][j] = ch[word[i][j]];
len[i] = j;
Insert(i);
}
}
int isequal(int x, int s)
{
int i;
for(i = 0; i < len[s]; i ++)
if(b[x + i] != code[s][i])
return 0;
return 1;
}
void dp(int x, int y)
{
int i, h = hash(b, x, y);
for(i = head[h]; i != -1; i = next[i])
if(len[i] == y - x + 1 && isequal(x, i))
break;
if(i != -1 && f[x - 1] + 1 < f[y])
{
f[y] = f[x - 1] + 1;
p[y] = x - 1, g[y] = i;
}
}
void print(int cur)
{
if(p[cur] == 0)
{
printf("%s", word[g[cur]]);
return ;
}
print(p[cur]);
printf(" %s", word[g[cur]]);
}
void solve()
{
int i, j, k;
memset(f, 0x3f, sizeof(f));
f[0] = 0;
for(i = 1; b[i]; i ++)
for(j = 1; j <= i; j ++)
dp(j, i);
-- i;
if(f[i] == INF)
printf("No solution.\n");
else
{
print(i);
printf("\n");
}
}
int main()
{
for(;;)
{
scanf("%s", b + 1);
if(b[1] == '-')
break;
init();
solve();
}
return 0;
}