链接：戳这里
A + B 就不写了吧
C. Reberland Linguistics time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word — some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction — it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways:
, where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.
Input
The only line contains a string s (5 ≤ |s| ≤ 104) consisting of lowercase English letters.
Output
On the first line print integer k — a number of distinct possible suffixes. On the next k lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Examples input
abacabaca
output
3
aca
ba
ca
input
abaca
output
Note
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
题意：
给出一个字符串,要求从后面开始分别连续的切（2|3）的字串下来,可以一直切到i>5的任意位置就算答案
这里连续的切长度为（2|3）的后缀字串还有一个要求,就是相邻的两个字串假如长度都为2,则这两个字串不能相等
同理长度为3也一样
要求输出能切的字串的个数,并把所有的字串字典序输出
比赛的时候我看错了一点条件,只是连续的两个长度相等的字串不能一样,而不是出现过就不能一样了(mdzz)
思路：
既然必须从后往前连续的分别去切长度为（2|3）的字串,那么我们把字符串翻转一下,设置dp状态
dp[i][0]表示第i个位置取前两个并且可取 dp[i][0]=1
如何判断可取 即 dp[i-2][1]=1 || (dp[i-2][0] && (s[i]!=s[i-2] || s[i-1]!=s[i-3]))
dp[i-2][1]=1表示第i-2的位置取了前三个
dp[i-2][0] && (s[i]!=s[i-2] || s[i-1]!=s[i-3]) 表示第i-2个位置取了前两个
并且当前的字串（s[i]s[i-1]）！=（s[i-2]s[i-3]）
dp[i][1]表示第i个位置取前三个并且可取 dp[i][1]=1
显然和上面的状态差不多
具体看代码
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int F[10100][2];
vector<string > V;
char s[10010];
int main(){
cin>>s;
int n=strlen(s);
if(n<=5) {
printf("0\n");
return 0;
}
reverse(s,s+n);
if(n-5>=2){string p="";p+=s[1];p+=s[0];V.push_back(p);}
if(n-5>=3){string p="";p+=s[2];p+=s[1];p+=s[0];V.push_back(p);}
F[1][0]=F[2][1]=1;
for(int i=3;i<n-5;i++){
if( F[i-2][1] || (F[i-2][0] && (s[i]!=s[i-2] || s[i-1]!=s[i-3])) ){
string p="";p+=s[i];p+=s[i-1];V.push_back(p);
F[i][0]=1;
}
if(i>=4){
if( F[i-3][0] || (i-5>=0 && F[i-3][1] && (s[i]!=s[i-3] || s[i-1]!=s[i-4] || s[i-2]!=s[i-5])) ){
string p="";p+=s[i];p+=s[i-1];p+=s[i-2];V.push_back(p);
F[i][1]=1;
}
}
}
sort(V.begin(),V.end());
V.erase(unique(V.begin(),V.end()),V.end());
cout<<V.size()<<endl;
for(int i=0;i<V.size();i++)
cout<<V[i]<<endl;
return 0;
}
D. World Tour time limit per test 5 seconds memory limit per test 512 megabytes input standard input output standard output
A famous sculptor Cicasso goes to a world tour!
Well, it is not actually a world-wide. But not everyone should have the opportunity to see works of sculptor, shouldn't he? Otherwise there will be no any exclusivity. So Cicasso will entirely hold the world tour in his native country — Berland.
Cicasso is very devoted to his work and he wants to be distracted as little as possible. Therefore he will visit only four cities. These cities will be different, so no one could think that he has "favourites". Of course, to save money, he will chose the shortest paths between these cities. But as you have probably guessed, Cicasso is a weird person. Although he doesn't like to organize exhibitions, he likes to travel around the country and enjoy its scenery. So he wants the total distance which he will travel to be as large as possible. However, the sculptor is bad in planning, so he asks you for help.
There are n cities and m one-way roads in Berland. You have to choose four different cities, which Cicasso will visit and also determine the order in which he will visit them. So that the total distance he will travel, if he visits cities in your order, starting from the first city in your list, and ending in the last, choosing each time the shortest route between a pair of cities — will be the largest.
Note that intermediate routes may pass through the cities, which are assigned to the tour, as well as pass twice through the same city. For example, the tour can look like that:
. Four cities in the order of visiting marked as overlines:[1, 5, 2, 4].
Note that Berland is a high-tech country. So using nanotechnologies all roads were altered so that they have the same length. For the same reason moving using regular cars is not very popular in the country, and it can happen that there are such pairs of cities, one of which generally can not be reached by car from the other one. However, Cicasso is very conservative and cannot travel without the car. Choose cities so that the sculptor can make the tour using only the automobile. It is guaranteed that it is always possible to do.
Input
In the first line there is a pair of integers n and m (4 ≤ n ≤ 3000, 3 ≤ m ≤ 5000) — a number of cities and one-way roads in Berland.
Each of the next m lines contains a pair of integers ui, vi (1 ≤ ui, vi ≤ n) — a one-way road from the city ui to the city vi. Note that uiand vi are not required to be distinct. Moreover, it can be several one-way roads between the same pair of cities.
Output
Print four integers — numbers of cities which Cicasso will visit according to optimal choice of the route. Numbers of cities should be printed in the order that Cicasso will visit them. If there are multiple solutions, print any of them.
Example input
8 9
1 2
2 3
3 4
4 1
4 5
5 6
6 7
7 8
8 5
output
2 1 8 7
Note
Let d(x, y) be the shortest distance between cities x and y. Then in the example d(2, 1) = 3, d(1, 8) = 7, d(8, 7) = 3.
The total distance equals 13.
题意：
给出n个点m条边的有向图, 每条边的长度为1
要求找出四个点a,b,c,d使得 a->b + b->c + c->d 的距离最大 其中a->b ,b->c ,c->d代表两点之间的最短路径
(1<=n<=3000) (1<=m<=5000)
思路：
1：先用复杂度内的最短路算法求出所有的i->j的最短路
2：我们可以先枚举所有的b->c 然后再去枚举a和d。显然对答案的贡献我们要求最大,所以a->b 和 c->的要尽可能的大
所以我们可以直接求出最大的a->b 和 c->d 但是因为a,b,c,d不能重点 ,我们直接枚举前三大的就可以了
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF 1e8
using namespace std;
int n,m,tot;
struct edge{
int v,w,next;
}e[5050];
int head[3010],vis[3010];
int mp[3030][3030],dis[3030];
vector< pair<int ,int> > V1[5050];
vector< pair<int ,int> > V2[5050];
void init(){
mst(head,-1);
tot=0;
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(i==j) mp[i][j]=0;
else mp[i][j]=INF;
}
}
}
void Add(int u,int v){
e[tot].v=v;
e[tot].w=1;
e[tot].next=head[u];
head[u]=tot++;
}
void SPFA(int x){
mst(vis,0);
for(int i=1;i<=n;i++) dis[i]=INF;
dis[x]=0;
vis[x]=1;
queue<int> qu;
qu.push(x);
while(!qu.empty()){
int u=qu.front();
qu.pop();
vis[u]=0;
for(int i=head[u];i!=-1;i=e[i].next){
int v=e[i].v;
if(dis[v]>dis[u]+e[i].w){
dis[v]=dis[u]+e[i].w;
if(!vis[v]){
vis[v]=1;
qu.push(v);
}
}
}
}
for(int i=1;i<=n;i++)
mp[x][i]=min(mp[x][i],dis[i]);
}
void solve(){
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(i==j) continue;
if(mp[i][j]!=INF) V1[i].push_back(make_pair(mp[i][j],j));
if(mp[j][i]!=INF) V2[i].push_back(make_pair(mp[j][i],j));
}
}
for(int i=1;i<=n;i++){
sort(V1[i].begin(),V1[i].end());
reverse(V1[i].begin(),V1[i].end());
sort(V2[i].begin(),V2[i].end());
reverse(V2[i].begin(),V2[i].end());
}
int t1,t2,t3,t4,ans=0;
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(i==j || mp[i][j]==INF) continue;
for(int k=0;k<3 && k<V1[j].size();k++){
if(i==V1[j][k].second) continue;
if(j==V1[j][k].second) continue;
for(int l=0;l<3 && l<V2[i].size();l++){
if(i==V2[i][l].second) continue;
if(j==V2[i][l].second) continue;
if(V1[j][k].second==V2[i][l].second) continue;
int tmp=V2[i][l].first + mp[i][j] + V1[j][k].first;
if(tmp > ans){
ans=tmp;
t1=V2[i][l].second;
t2=i;
t3=j;
t4=V1[j][k].second;
}
}
}
}
}
cout<<t1<<" "<<t2<<" "<<t3<<" "<<t4<<endl;
///cout<<ans<<endl;
}
int main(){
scanf("%d%d",&n,&m);
init();
for(int i=1;i<=m;i++){
int u,v;
scanf("%d%d",&u,&v);
Add(u,v);
}
for(int i=1;i<=n;i++) SPFA(i);
/*for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
printf("%d ",mp[i][j]);
}
cout<<endl;
}*/
solve();
return 0;
}