Bless You Autocorrect!（字典树上建图）

传送门：QAQ
题意：实现一个单词补全功能的最小操作步数

思路：对于读入的单词表，我们可以预处理出到达每个前缀所需要的最小步数，这我们可以实现建立一颗字典树，然后在字典树上跑bfs即可。

预处理完后，我们对于读入的单词，只要遍历它的每个前缀统计最小次数即可。

 

附上代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<map>
#include<queue>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
const int MAX_NODE = 1100000 + 10;
const int MAXM=1e5+10;
const int CHARSET = 26;
int trie[MAX_NODE][CHARSET];
char ch[MAX_NODE];
int CharCnt[MAX_NODE];
int vis[MAX_NODE];
int num[MAX_NODE];
int cnt = 1;
vector<int>gx[MAX_NODE];
void insert(char *w) {
	int len = strlen(w);
	int p = 0;
	for (int i = 0; i < len; i++) {
		int c = w[i] - 'a';
		if (!trie[p][c]) {
			trie[p][c] = cnt;
			CharCnt[cnt] = 0;
			cnt++;
		}
		CharCnt[trie[p][c]]++;
		gx[p].push_back(trie[p][c]);
		gx[trie[p][c]].push_back(p);
		p = trie[p][c];
	}
	int id = 0;
	for (int i = 0; i < len; i++) {
		int c = w[i] - 'a';
		if (CharCnt[trie[id][c]] == 1) {
			gx[p].push_back(trie[id][c]);
			gx[trie[id][c]].push_back(p);
			break;
		}
		id = trie[id][c];
	}
}
void bfs() {
	queue<int>q;
	q.push(0);
	num[0] = 0;
	vis[0] = 1;
	while (!q.empty()) {
		int st = q.front();
		vis[st] = 1;
		q.pop();
		for (int i = 0; i < gx[st].size(); i++) {
			if (vis[gx[st][i]]) continue;
			if (num[gx[st][i]] > num[st] + 1) {
				num[gx[st][i]] = num[st] + 1;
				q.push(gx[st][i]);
			}
		}
	}
}
void init() {
	cnt = 1;
	memset(trie, 0, sizeof(trie));
	memset(vis, 0, sizeof(vis));
	for (int i = 0; i < MAX_NODE; i++) {
		num[i] = inf;
	}
}
int main(void) {
	int n, m;
	init();
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++) {
		scanf("%s", ch);
		insert(ch);
	}
	bfs();
	for (int i = 0; i < m; i++) {
		scanf("%s", ch);
		int len = strlen(ch);
		int id = 0, min_ans = len;
		for (int i = 0; i < len; i++) {
			int w = ch[i] - 'a';
			if (!trie[id][w]) break;
			id = trie[id][w];
			min_ans = min(min_ans, num[id] + len-1 - i);
			//printf("%c**%d**\n", ch[i], num[id]);
		}
		printf("%d\n", min_ans);
	}
}