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;
}