Problem Solving►UVa

UVa 429 - Word Transformation

Contents

  1. 1. Problem
  2. 2. Solution
  3. 3. Code

Problem

題目網址
中文網址

Solution

將所有詞建成一張圖，兩個詞的 漢明距離 若等於 1 就將它們相連在一起。
接著以開始字當起點做 BFS 即可。

Code

UVa 429
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#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#define N 201
#define MIN(a, b) ((a) < (b) ? (a) : (b))
using namespace std;

int main()
{
    int T;
    char str[N][15];
    vector<int> adj_list[N];
    scanf("%d ", &T);
    while (T--)
    {

        int idx = 1;
        while (fgets(str[idx], 15, stdin) && str[idx][0] != '*')
        {
            str[idx][strlen(str[idx]) - 1] = '\0';
            ++idx;
        }

        for (int i = 1; i < idx; ++i)
        {
            for (int j = i + 1; j < idx; ++j)
            {
                int diff = 0;
                for (int k = 0; diff < 2 && str[i][k] && str[j][k]; ++k)
                    if (str[i][k] != str[j][k])
                        ++diff;

                if (diff < 2)
                {
                    adj_list[i].push_back(j);
                    adj_list[j].push_back(i);
                }                
            }
        }

        char tmp[30], a[15], b[15];
        int step[N] = {};
        while (fgets(tmp, 30, stdin) && tmp[0] != '\n')
        {
            int i, j, a_id, b_id;

            for (i = 0; tmp[i] != ' '; ++i)
                a[i] = tmp[i];
            a[i++] = '\0';
            for (j = 0; tmp[i] != '\n'; ++i, ++j)
                b[j] = tmp[i];
            b[j] = '\0';

            for (i = 1; i < idx; ++i)
                if (!strcmp(a, str[i]))
                {
                    a_id = i;
                    break;
                }

            for (i = 1; i < idx; ++i)
                if (!strcmp(b, str[i]))
                {
                    b_id = i;
                    break;
                }

            //BFS
            queue<int> Q;
            Q.push(a_id);
            step[a_id] = 1;
            while (!Q.empty() && !step[b_id])
            {
                int now = Q.front();
                Q.pop();
                for (int i : adj_list[now])
                {
                    if (!step[i])
                    {
                        step[i] = step[now] + 1;
                        if (i == b_id)
                            break;
                        Q.push(i);
                    }
                }
            }

            printf("%s %s %d\n", a, b, step[b_id] - 1);
            memset(step, 0, sizeof step);
        }

        if (T)
            putchar('\n');

        for (int i = 0; i < idx; ++i)
            adj_list[i].clear();
    }

    return 0;
}