Codeforces Round #349 (Div. 2) - C
来源:互联网 发布:js url encode php 编辑:程序博客网 时间:2024/05/22 12:59
题目链接:http://codeforces.com/contest/667/problem/C
参考博客:http://www.cnblogs.com/qscqesze/p/5448085.html
题意:
给一个字符串,然后你你需要切一个长度至少为5的前缀下来,然后剩下的都得切成是长度为2或者3的字符串
你需要连续的切出来的字符串都不一样,问你能够切出多少不同的块
思路:晚上做的时候,英语硬伤,题意理解成,后缀中不能出现相同的部分,就以为是dfs,然后不是WA就是TLE,网上搜题解,看了大神的思路用dp做
#include<iostream>#include<cstring>#include<algorithm>#include<set>#include<cstdio>#include<cstdlib>#include<string>#include<map>#include<cstdio>#include<cstdlib>#include<vector>#include<cmath>#include<queue>#include<list>#define PI acos(-1.0)#define esp 1e-12using namespace std;typedef long long ll;typedef pair<int,int> P;set<string> s;string temp_2, temp_3;string tt_2, tt_3;char str[10000+50];int dp[10000+50][2]={0};bool vis_2[30][30];//bool vis_3[30][30][30];int main(){//memset(vis_2,0,sizeof(vis_2));memset(vis_3,0,sizeof(vis_3));cin>>str;int n = strlen(str);reverse(str,str+n);dp[1][0]=1; dp[2][1] = 1;if(n-5>=2){tt_2.clear();tt_2.push_back(str[1]); tt_2.push_back(str[0]);s.insert(tt_2);//vis_2[str[1]-'a'][str[0]-'a']=1;temp_2=tt_2;}if(n-5>=3){tt_3.clear();tt_3.push_back(str[2]); tt_3.push_back(str[1]);tt_3.push_back(str[0]);s.insert(tt_3);//vis_3[str[2]-'a'][str[1]-'a'][str[0]-'a'] = 1;temp_3=tt_3;}for(int i = 3; i < n-5; i++){if(i-2>=0){if(dp[i-2][1]){tt_2.clear();tt_2.push_back(str[i]);tt_2.push_back(str[i-1]);//if(!vis_2[str[i]-'a'][str[i-1]-'a'])s.insert(tt_2);temp_2 = tt_2;dp[i][0] = 1;}else if(dp[i-2][0]){tt_2.clear();tt_2.push_back(str[i]);tt_2.push_back(str[i-1]);if(tt_2!=temp_2){//if(!vis_2[str[i]-'a'][str[i-1]-'a'])s.insert(tt_2);temp_2 = tt_2;dp[i][0] = 1;}}}if(i-3>=0){if(dp[i-3][0]){tt_3.clear();tt_3.push_back(str[i]);tt_3.push_back(str[i-1]);tt_3.push_back(str[i-2]);//if(!vis_3[str[i]-'a'][str[i-1]-'a'][str[i]-'a'])s.insert(tt_3);temp_3 = tt_3; dp[i][1] = 1;}else if(dp[i-3][1]){tt_3.clear();tt_3.push_back(str[i]);tt_3.push_back(str[i-1]);tt_3.push_back(str[i-2]);if(tt_3!=temp_3){//if(!vis_3[str[i]-'a'][str[i-1]-'a'][str[i]-'a'])s.insert(tt_3);temp_3=tt_3;dp[i][1] = 1;}}}}set<string>::iterator it;cout<<s.size()<<endl;for(it = s.begin(); it != s.end(); it++)cout<<*it<<endl;return 0;}
0 0
- Codeforces Round #349 (Div. 2) - C
- 【codeforces】Codeforces Round #370 (Div. 2) C
- Codeforces Round #349 (Div. 2)
- Codeforces Round #349 (Div. 2) C. Reberland Linguistics
- Codeforces Round #349 (Div. 2) C. Reberland Linguistics 【DP】
- Codeforces Round #349 (Div. 2) C. Reberland Linguistics DP
- Codeforces Round #105 (Div. 2) C
- Codeforces Round 134 div 2 C题
- Codeforces Round #137 (Div. 2), problem: (C)
- Codeforces Round #153 (Div. 2) C题
- Codeforces Round #158 (Div. 2) C题
- Codeforces Round #162 (Div. 2) C
- Codeforces Round #166 (Div. 2) c. Secret
- Codeforces Round#170(Div 2)C
- Codeforces Round #173 (Div. 2) Problem C
- Codeforces Round #192 (Div. 2) C. Purification
- Codeforces Round #196 (Div. 2) C. Quiz
- Codeforces Round #197 (Div. 2) (C~E)
- 离散数学之集合上关系的判断
- [离散化]图形面积
- Boost Test学习总结(C++)
- poj 2777 Count Color(线段树区间更新+位运算)
- mesos上运行docker registry
- Codeforces Round #349 (Div. 2) - C
- 发表在IBM Developworks上的文章,Spark Streaming 图片处理案例介绍
- [乱搞 暴力 并查集] BZOJ 3563 DZY Loves Chinese
- maven在不同环境下打包的应用
- Java 类加载机制详解
- [三进制倍增 || 可并堆] BZOJ 4003 [JLOI2015]城池攻占
- Java的对象是采用值传递还是引用传递?
- SqlServer数据库同步 两张表的数据 去除重复数据
- Java重载(OverLoad)的理解