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who is the best?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Problem Description
There are N people want to choose the best person. Each person select the best person ai, .John wants to know that who received the most number of votes.
Input
The first line contains a single integer T(1≤T≤50),indicating the number of test cases.
Each test case begins with an integer N(1≤N≤100),indicating the number of person.
Next N lines contains an integer ai(1≤ai≤N).
Each test case begins with an integer N(1≤N≤100),indicating the number of person.
Next N lines contains an integer ai(1≤ai≤N).
Output
For each case, output an integer means who is the best person. If there are multiple answers, print the minimum index.
Sample Input
2
10
1
2
3
4
5
6
7
8
9
10
2
3
4
5
6
7
8
9
10
5
3
3
3
3
3
3
3
3
3
Sample Output
1
3
3
题意：求出现次数最多的数，若有多个数，则输出最小的一个
水题，随便搞
#include<iostream>
#include <cstring>
using namespace std;
int a[];
int main()
{
ios::sync_with_stdio(false);
int t;
cin>>t;
int n;
memset(a,,sizeof(a));
while(t--)
{
int n;
cin>>n;
int k;
int maxx=;
memset(a,,sizeof(a));
int ans=;
for(int i=;i<n;i++)
{
cin>>k;
a[k]++;
if(a[k]>=maxx)
{
if(a[k]==maxx)
{
ans=min(ans,k);
}
else
{
ans=k;
}
maxx=a[k];
}
}
cout<<ans<<endl;
}
return ;
}
代码君
lines
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Problem Description
John
has several lines. The lines are covered on the X axis. Let A is a
point which is covered by the most lines. John wants to know how many
lines cover A.
has several lines. The lines are covered on the X axis. Let A is a
point which is covered by the most lines. John wants to know how many
lines cover A.
Input
The first line contains a single integer T(1≤T≤100)(the data for N>100 less than 11 cases),indicating the number of test cases.
Each test case begins with an integer N(1≤N≤105),indicating the number of lines.
Next N lines contains two integers Xi and Yi(1≤Xi≤Yi≤109),describing a line.
Each test case begins with an integer N(1≤N≤105),indicating the number of lines.
Next N lines contains two integers Xi and Yi(1≤Xi≤Yi≤109),describing a line.
Output
For each case, output an integer means how many lines cover A.
Sample Input
2
5
1 2
2 2
2 4
3 4
5 1000
5
1 1
2 2
3 3
4 4
5 5
Sample Output
3
1
1
题意：有n条线段，求被覆盖到次数最多的点的次数
分析：
1.可以转化成求前缀和最大的问题：将区间改成左闭右开（即右端点加1），排序，从左往右遍历，若为左端点则加一，右端点则减一。
2.线段树，离散化一下，然后区间更新，单点查询。
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef pair<int,int> PII;
PII a[];
int main()
{
ios::sync_with_stdio(false);
int t;
//freopen("in.in","r",stdin);
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
n=n*;
for(int i=;i<n;i++)
{
scanf("%d",&a[i].first);
a[i].second=;
scanf("%d",&a[++i].first);
a[i].first++;
a[i].second=-;
}
sort(a,a+n);
int ans=;
int k=;
for(int i=;i<n;i++)
{
k=k+a[i].second;
ans=max(k,ans);
}
printf("%d\n",ans);
}
}
代码君
magic balls
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Problem Description
The town of W has N people. Each person takes two magic balls A and B every day. Each ball has the volume ai and bi.
People often stand together. The wizard will find the longest
increasing subsequence in the ball A. The wizard has M energy. Each
point of energy can change the two balls’ volume.(swap(ai,bi)).The
wizard wants to know how to make the longest increasing subsequence and
the energy is not negative in last. In order to simplify the problem,
you only need to output how long the longest increasing subsequence is.
People often stand together. The wizard will find the longest
increasing subsequence in the ball A. The wizard has M energy. Each
point of energy can change the two balls’ volume.(swap(ai,bi)).The
wizard wants to know how to make the longest increasing subsequence and
the energy is not negative in last. In order to simplify the problem,
you only need to output how long the longest increasing subsequence is.
Input
The first line contains a single integer T(1≤T≤20)(the data for N>100 less than 6 cases), indicating the number of test cases.
Each test case begins with two integer N(1≤N≤1000) and M(0≤M≤1000),indicating the number of people and the number of the wizard’s energy. Next N lines contains two integer ai and bi(1≤ai,bi≤109),indicating the balls’ volume.
Each test case begins with two integer N(1≤N≤1000) and M(0≤M≤1000),indicating the number of people and the number of the wizard’s energy. Next N lines contains two integer ai and bi(1≤ai,bi≤109),indicating the balls’ volume.
Output
For each case, output an integer means how long the longest increasing subsequence is.
Sample Input
2
5 3
5 1
4 2
3 1
2 4
3 1
5 4
5 1
4 2
3 1
2 4
3 1
Sample Output
4
4
题意：有两组数a,b，另外最多可以对其做m次操作——即swap(a[i],b[i])，问在此情况下a数组最大的LIS长度是多少？
分析：先将数据离散化，然后用m+1个树状数组或者线段树来维护在还剩余j次操作之后到第i个数时的最大的LIS长度。
sad，这道题我用了线段树来维护RMQ，一直TLE，最后在参考了邝巨巨的情况下才过了的。
^_^通过这道题目知道了如何运用树状数组来维护RMQ，之前只会用线段树搞搞。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
struct BIT
{
int bit[];
int n;
int init(int size)
{
n=size;
for(int i=;i<n;i++)bit[i]=;
}
int query(int i)
{
int s=;
while(i>)
{
s=max(s,bit[i]);
i-=i&-i;
}
return s;
}
int update(int i,int x)
{
while(i<=n)
{
bit[i]=max(x,bit[i]);
i+=i&-i;
}
}
}bt[];
int a[],b[],c[];
int main()
{
//ios::sync_with_stdio(false);
int t;
//freopen("in.in","r",stdin);
scanf("%d",&t);
while(t--)
{
int n,m;
scanf("%d%d",&n,&m);
int cnt=;
for(int i=;i<n;i++)
{
scanf("%d%d",&a[i],&b[i]);
c[++cnt]=a[i];
c[++cnt]=b[i];
}
sort(c+,c+cnt+);
cnt=unique(c+,c+cnt+)-c-;
for(int i=;i<n;i++)
{
a[i]=lower_bound(c+,c+cnt+,a[i])-c;
b[i]=lower_bound(c+,c+cnt+,b[i])-c;
}
int ans=;
for(int i=;i<=m;i++)bt[i].init(cnt);
for(int i=;i<n;i++)
{
for(int j=;j<=m;j++)
{
int x=bt[j].query(a[i]-)+;
bt[j].update(a[i],x);
ans=max(ans,x);
if(j<m)
{
x=bt[j+].query(b[i]-)+;
bt[j].update(b[i],x);
ans=max(ans,x);
}
}
}
cout<<ans<<endl; } return ;
}
代码君
D题
目前还不会，不会cdq分治。。。