CodeForces Round #403 补题

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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 100005int 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;}

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