Word Search

题意：单词搜索，在一个单词网中搜索单词，该单词可能存在这个网的水平，垂直，或对角线方向。

no solution：有词汇未在方阵图中出现；

ambiguous：词汇在方阵图中出现了两次；

empty solution：所有词汇均出现且仅出现一次，最终方阵图无未标记字母；

此外则依次输出剩余大写字母。

// Solution to Word Search
// Author: Thomas Beuman// Time complexity: O(n*h*w*wordlength)
// Memory: O(h*w)// @EXPECTED_RESULTS@: CORRECT#include <cstdio>
#include <cstring>
#include <string>
using namespace std;const int hmax = 32, wmax = 32, lmax = 32;char Grid[hmax][wmax+1];
char Word[lmax+1];bool Marked[hmax][wmax];
//控制8个方向矩阵
int di[] = {-1,-1,-1,0,0,1,1,1};
int dj[] = {-1,0,1,-1,1,-1,0,1};int main()
{   int runs, run, n, h, w, d, i, j, k, m, s, i2, j2, solutions;bool nosolution, ambiguous, palindrome, empty;scanf("%d", &runs);for (run = 0; run < runs; run++){  // Read inputscanf("%d %d %d", &n, &h, &w);for (i = 0; i < h; i++)scanf("%s", Grid[i]);memset(Marked, false, sizeof(Marked));nosolution = ambiguous = false;// Process the wordsfor (s = 0; s < n; s++){  scanf("%s", Word);m = strlen(Word);//检查是否回文for (k = 0; k < (m-1)/2 && Word[k] == Word[m-1-k]; k++);palindrome = (k == (m-1)/2);solutions = 0;// 尝试所有的开始的位置和矩阵for (i = 0; i < h; i++)for (j = 0; j < w; j++)for (d = 0; d < 8; d++){    for (k = 0; k < m; k++){  i2 = i + k * di[d];j2 = j + k * dj[d];if (i2 < 0 || i2 >= h || j2 < 0 || j2 >= w || Grid[i2][j2] != Word[k])break;}if (k == m) // 查找到单词{   for (k = 0; k < m; k++)Marked[i+k*di[d]][j+k*dj[d]] = true;solutions++;}}if (solutions == 0)nosolution = true;if (!(solutions == 1 || (palindrome && solutions == 2) || (m == 1 && solutions == 8)))ambiguous = true;}// 输出答案if (nosolution)printf("no solution\n");else if (ambiguous)printf("ambiguous\n");else{   // Print unmarked lettersempty = true;for (i = 0; i < h; i++)for (j = 0; j < w; j++)if (!Marked[i][j]){    printf("%c", Grid[i][j]);empty = false;}if (empty)printf("empty solution");printf("\n");}}return 0;
}