Problem A :

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;

#define F(i,a,b) for(int i=a;i<=b;i++)

#define maxn 100005

int main(){
    int n,num[2*maxn];
    scanf("%d",&n);

    F(i,1,2*n) scanf("%d",&num[i]);

    int count=0,res=0,vis[maxn];
    memset(vis, 0, sizeof(vis));

    F(i,1,2*n){
        if(vis[num[i]]==0){
            vis[num[i]]=1;
            count++;
        }else{
            count--;
        }
        res = max(res, count);
    }
    cout<<res<<endl;
    return 0;
}

Problem B :

这题刚开始没有什么思路，后来想通了是用二分查找，注意在STL库里面max 和 min比较double，尾数越多比较的时间越长，选择 1e-7 才不容易TLE

#include <iostream>
#include <cstdio>
using namespace std;

const double eps=1e-7;

#define maxn 60005
#define F(i,a,b) for(int i=a;i<=b;i++)

int n,x[maxn],v[maxn],sum;

bool check(double m){
    double b=-2e9,t=2e9;
    F(i,1,n){
        b = max(b,x[i]-v[i]*m);
        t = min(t,x[i]+v[i]*m);
    }
    return b<=t;
}

int main(){

    sum =0;

    scanf("%d",&n);
    F(i,1,n) {scanf("%d",&x[i]);sum+=x[i];}
    F(i,1,n) scanf("%d",&v[i]);

    double L=0,R=1e9,M;

    while((R-L)>=eps){
        M = (L+R)/2.0;
        if(check(M)) R=M;
        else L=M;
    }
    printf("%.9lf\n",(L+R)/2.0);

    return 0;
}

Problem C :

#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int  maxn = 2e5+5;


int n,color[maxn],vis[maxn],res=0;
vector<int> v[maxn];

void dfs(int d){
    int now = 1;
    vector<int>::iterator it;
    for(it=v[d].begin();it!=v[d].end();it++){
        if(*it==vis[d]) continue;
        vis[*it]=d;
        while (color[d]==now || color[vis[d]]==now) now++;
        color[*it]=now;
        now++;
    }
    res=max(now-1,res);

    for(it=v[d].begin();it!=v[d].end();it++){
        if(*it!=vis[d])
            dfs(*it);
    }
}

int main(){
    scanf("%d",&n);
    for(int i=0,x,y;i<n-1;i++){
        scanf("%d%d",&x,&y);
        v[x].push_back(y);
        v[y].push_back(x);
    }

    color[1] = 1;

    dfs(1);

    cout<<res<<endl;

    for(int i=1;i<n;i++)
        printf("%d ",color[i]);
    cout<<color[n]<<endl;
    return 0;
}