Codeforces 667C Reberland Linguistics 记忆化搜索

链接

Codeforces 667C Reberland Linguistics

题意

给你一个字符串，除去前5个字符串后，使剩下的串有长度为2或3的词根组成，相邻的词根不能重复。找到所有的词根

思路

去掉前5个字符，将剩下的串反过来进行记忆化，用vis[last][pos]记录一下当前状态是否做过。last是之前与之相邻的词根。比赛的时候只用了vis[i]错了。

代码

#include <iostream>
#include <cstdio>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <map>
#include <set>
#include <cmath>
#include <cstring>
#include <string>

#define LL long long
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN 10005
using namespace std;

string s, ss;
int len;
vector<string> res;
map<string, bool> mp;
map<pair<string, int>, bool> vis;
int f[MAXN];
void dfs(int pos, string last){
	if (vis[make_pair(last, pos)] == true) return;
	int cur = -1;
	if (pos >= len) return;
	for (int i = 0; i < 2; ++i){
		if (pos + 2  + i<= len){
			string temp = s.substr(pos, 2 + i);
			if (!mp[temp]){
				res.push_back(temp);
				mp[temp] = true;
			}
			if (temp != last){
				dfs(pos + 2 + i, temp);
			}
		}
	}
	vis[make_pair(last,pos)] = true;
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
	cin >> ss;
	ss = ss.substr(5);
	len = ss.length();
	for (int i = len - 1; i >= 0; --i){
		s.push_back(ss[i]);
	}
	dfs(0, "");
	int ans = res.size();
	printf("%d\n", ans);
	for (int i = 0; i < res.size(); ++i){
		int len = res[i].length();
		for (int j = 0; j < (len >> 1); ++j){
			swap(res[i][j], res[i][len - j - 1]);
		}
	}
	sort(res.begin(), res.end());
	for (int i = 0; i < res.size(); ++i){
		cout << res[i] << endl;
	}
}