我的代码应该不会被hack，立个flag
1 second
256 megabytes
standard input
standard output
It is winter now, and Max decided it's about time he watered the garden.
The garden can be represented as n consecutive garden beds, numbered from 1 to n. k beds contain water taps (i-th tap is located in the bed xi), which, if turned on, start delivering water to neighbouring beds. If the tap on the bed xi is turned on, then after one second has passed, the bed xi will be watered; after two seconds have passed, the beds from the segment [xi - 1, xi + 1] will be watered (if they exist); after j seconds have passed (j is an integer number), the beds from the segment [xi - (j - 1), xi + (j - 1)] will be watered (if they exist). Nothing changes during the seconds, so, for example, we can't say that the segment [xi - 2.5, xi + 2.5] will be watered after 2.5seconds have passed; only the segment [xi - 2, xi + 2] will be watered at that moment.
The garden from test 1. White colour denotes a garden bed without a tap, red colour — a garden bed with a tap.
The garden from test 1 after 2 seconds have passed after turning on the tap. White colour denotes an unwatered garden bed, blue colour — a watered bed.
Max wants to turn on all the water taps at the same moment, and now he wonders, what is the minimum number of seconds that have to pass after he turns on some taps until the whole garden is watered. Help him to find the answer!
The first line contains one integer t — the number of test cases to solve (1 ≤ t ≤ 200).
Then t test cases follow. The first line of each test case contains two integers n and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n) — the number of garden beds and water taps, respectively.
Next line contains k integers xi (1 ≤ xi ≤ n) — the location of i-th water tap. It is guaranteed that for each 
condition xi - 1 < xiholds.
It is guaranteed that the sum of n over all test cases doesn't exceed 200.
Note that in hacks you have to set t = 1.
For each test case print one integer — the minimum number of seconds that have to pass after Max turns on some of the water taps, until the whole garden is watered.
3
5 1
3
3 3
1 2 3
4 1
1
3
1
4
The first example consists of 3 tests:
- There are 5 garden beds, and a water tap in the bed 3. If we turn it on, then after 1 second passes, only bed 3 will be watered; after 2seconds pass, beds [1, 3] will be watered, and after 3 seconds pass, everything will be watered.
- There are 3 garden beds, and there is a water tap in each one. If we turn all of them on, then everything will be watered after 1 second passes.
- There are 4 garden beds, and only one tap in the bed 1. It will take 4 seconds to water, for example, bed 4.
A题意很长是个模拟，自己也迷了好久，枚举每个点就可以了
#include<bits/stdc++.h>
using namespace std;
int a[],b[];
int main()
{
ios::sync_with_stdio(false);
int t;
cin>>t;
while(t--)
{
memset(b,,sizeof(b));
int n,k,ma=;
cin>>n>>k;
for(int i=; i<k; i++)
cin>>a[i],b[a[i]]=;
for(int i=; i<=n; i++)
{
int sum=,tl=i,tr=i;
while(b[tl]!=&&b[tr]!=)
{
if(tl->)tl--;
if(tr+<=n)tr++;
sum++;
}
ma=max(ma,sum);
}
cout<<ma<<endl;
}
return ;
}
qls的代码超级优秀的，妈耶，长知识
#include<bits/stdc++.h>
using namespace std;
const int MAXN=;
int x[MAXN];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,k;
scanf("%d%d",&n,&k);
for(int i=;i<=k;i++)
scanf("%d",&x[i]);
int res=;
for(int i=;i<=n;i++)
{
int mi=n;
for(int j=;j<=k;j++)
mi=min(mi,abs(i-x[j]));
res=max(res,mi);
}
printf("%d\n",res+);
}
return ;
}
1 second
256 megabytes
standard input
standard output
Recently n students from city S moved to city P to attend a programming camp.
They moved there by train. In the evening, all students in the train decided that they want to drink some tea. Of course, no two people can use the same teapot simultaneously, so the students had to form a queue to get their tea.
i-th student comes to the end of the queue at the beginning of li-th second. If there are multiple students coming to the queue in the same moment, then the student with greater index comes after the student with lesser index. Students in the queue behave as follows: if there is nobody in the queue before the student, then he uses the teapot for exactly one second and leaves the queue with his tea; otherwise the student waits for the people before him to get their tea. If at the beginning of ri-th second student i still cannot get his tea (there is someone before him in the queue), then he leaves the queue without getting any tea.
For each student determine the second he will use the teapot and get his tea (if he actually gets it).
The first line contains one integer t — the number of test cases to solve (1 ≤ t ≤ 1000).
Then t test cases follow. The first line of each test case contains one integer n (1 ≤ n ≤ 1000) — the number of students.
Then n lines follow. Each line contains two integer li, ri (1 ≤ li ≤ ri ≤ 5000) — the second i-th student comes to the end of the queue, and the second he leaves the queue if he still cannot get his tea.
It is guaranteed that for every 
condition li - 1 ≤ li holds.
The sum of n over all test cases doesn't exceed 1000.
Note that in hacks you have to set t = 1.
For each test case print n integers. i-th of them must be equal to the second when i-th student gets his tea, or 0 if he leaves without tea.
2
2
1 3
1 4
3
1 5
1 1
2 3
1 2
1 0 2
The example contains 2 tests:
- During 1-st second, students 1 and 2 come to the queue, and student 1 gets his tea. Student 2 gets his tea during 2-nd second.
- During 1-st second, students 1 and 2 come to the queue, student 1 gets his tea, and student 2 leaves without tea. During 2-nd second, student 3 comes and gets his tea.
n个人排队，有进队时间和出队时间，进队时间相同的话，编号小的优先
队列或者数组模拟吧
#include<bits/stdc++.h>
using namespace std;
int l[],r[],M[];
int main()
{
ios::sync_with_stdio(false);
int T;
cin>>T;
while(T--)
{
int n;
cin>>n;
memset(M,,sizeof M);
for(int i=; i<=n; i++)
cin>>l[i]>>r[i];
int tot=;
queue<int> q;
for(int i=; i<=; i++)
{
while(tot<=n&&l[tot]==i)
q.push(tot++);
while(q.size()&&r[q.front()]<i)
q.pop();
if(q.size())
{
int tmp=q.front();
q.pop();
M[tmp]=i;
}
}
for(int i=; i<=n; i++)
cout<<M[i]<<" ";
cout<<"\n";
}
return ;
}
vector的
#include<bits/stdc++.h>
using namespace std;
vector<pair<int,pair<int,int> > >V;
int M[];
int main()
{
ios::sync_with_stdio(false);
int T;
cin>>T;
while(T--)
{
V.clear();
int n;
cin>>n;
memset(M,,sizeof M);
for(int i=,l,r; i<=n; i++)
cin>>l>>r,V.push_back(make_pair(l,make_pair(i,r)));
sort(V.begin(),V.end());
int j=,l=n;
for(int i=; i<= &&j<n; i++)
{
while((V[j].second).second<i&&j<n)j++;
if(V[j].first<=i)
M[(V[j].second).first]=i,j++;
}
for(int i=; i<=n; i++)
cout<<M[i]<<" ";
cout<<"\n";
}
return ;
}
qls很优秀的代码
#include<bits/stdc++.h>
using namespace std;
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
int now=;
for(int i=;i<=n;i++)
{
int l,r;
scanf("%d%d",&l,&r);
if(now>r)printf("");
else
{
now=max(now,l);
printf("%d",now++);
}
printf("%c"," \n"[i==n]);
}
}
return ;
}
1 second
256 megabytes
standard input
standard output
You have an array a consisting of n integers. Each integer from 1 to n appears exactly once in this array.
For some indices i (1 ≤ i ≤ n - 1) it is possible to swap i-th element with (i + 1)-th, for other indices it is not possible. You may perform any number of swapping operations any order. There is no limit on the number of times you swap i-th element with (i + 1)-th (if the position is not forbidden).
Can you make this array sorted in ascending order performing some sequence of swapping operations?
The first line contains one integer n (2 ≤ n ≤ 200000) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 200000) — the elements of the array. Each integer from 1 to n appears exactly once.
The third line contains a string of n - 1 characters, each character is either 0 or 1. If i-th character is 1, then you can swap i-th element with (i + 1)-th any number of times, otherwise it is forbidden to swap i-th element with (i + 1)-th.
If it is possible to sort the array in ascending order using any sequence of swaps you are allowed to make, print YES. Otherwise, print NO.
6
1 2 5 3 4 6
01110
YES
6
1 2 5 3 4 6
01010
NO
In the first example you may swap a3 and a4, and then swap a4 and a5.
交换相邻的其实就是在排序啊，所以出现连续的1就去sort好了
#include<bits/stdc++.h>
using namespace std;
int a[];
int main()
{
int n;
while(cin>>n)
{
int f=;
string s;
for(int i=;i<=n;i++)
cin>>a[i];
cin>>s;
ios::sync_with_stdio(false);
int l=,r=;
for(int i=;s[i];i++)
{
if(s[i]=='')continue;
l=i;
for(;s[i];i++)
{
if(s[i]=='')break;
r=i;
}
sort(a+l+,a+r+);
}
for(int i=;i<=n&&f;i++)
if(a[i]!=i)f=;
if(f)cout<<"YES\n";
else cout<<"NO\n";
}
return ;
}
2 seconds
256 megabytes
standard input
standard output
You are given an undirected graph consisting of n vertices and 
edges. Instead of giving you the edges that exist in the graph, we give you m unordered pairs (x, y) such that there is no edge between x and y, and if some pair of vertices is not listed in the input, then there is an edge between these vertices.
You have to find the number of connected components in the graph and the size of each component. A connected component is a set of vertices X such that for every two vertices from this set there exists at least one path in the graph connecting these vertices, but adding any other vertex to X violates this rule.
The first line contains two integers n and m (1 ≤ n ≤ 200000, 
).
Then m lines follow, each containing a pair of integers x and y (1 ≤ x, y ≤ n, x ≠ y) denoting that there is no edge between x and y. Each pair is listed at most once; (x, y) and (y, x) are considered the same (so they are never listed in the same test). If some pair of vertices is not listed in the input, then there exists an edge between those vertices.
Firstly print k — the number of connected components in this graph.
Then print k integers — the sizes of components. You should output these integers in non-descending order.
5 5
1 2
3 4
3 2
4 2
2 5
2
1 4
#include<bits/stdc++.h>
using namespace std;
const int N=2e5+;
set<int>M[N+N];
queue<int>Q;
int q[N],yy[N],n,m,k;
void bfs()
{
while(!Q.empty())
{
int tot=,u=Q.front();
Q.pop();
q[]=u;
for(int i=; i<tot; i++)
{
int now=n-tot;
while(now>&&!Q.empty())
{
int v=Q.front();
Q.pop();
if(!M[q[i]].count(v)) q[tot++]=v;
else Q.push(v);
--now;
}
}
yy[k++]=tot;
}
}
int main()
{
ios::sync_with_stdio(false);
cin>>n>>m;
for(int i=,u,v; i<m; i++)
cin>>u>>v,M[u].insert(v),M[v].insert(u);
for(int i=; i<=n; i++) Q.push(i);
bfs();
sort(yy,yy+k);
printf("%d\n",k);
for(int i=; i<k; i++)printf("%d ",yy[i]);
return ;
}