一、一本通 1341 一笔画问题

1.1 问题描述

1.2 问题解决

#include<iostream>

using namespace std;
int n,m;
int g[N][N];
int dis[N],path[N*2];
int cnt;
void dfs(int i)
{
    for(int j=1;j<=n;j++)
        if(g[i][j])
        {
            g[i][j]=0;
            g[j][i]=0;
            dfs(j);
        }
    path[cnt++]=i;
}
int main()
{
    cin>>n>>m;
    for(int i=1;i<=m;i++)
    {
        int x,y;
        cin>>x>>y;
        g[x][y]=1;
        g[y][x]=1;
        dis[x]++;
        dis[y]++;
    }
 
    int start=1;
    int sum=0;
    for(int i=1;i<=n;i++)
        if(dis[i]%2)
        {
            sum++;
            if(sum==1)
                start=i;
        }
 
    dfs(start);
 
    for(int i=cnt-1;i>=0;i--)
        cout<<path[i]<<" ";
    return 0;
}

二、一本通 1374 铲雪车

2.1 问题描述

2.2 问题解决

#include<iostream>

using namespace std;
int main()
{
    long long start,endd;
    long long x1,y1,x2,y2;
    double dis=0;
 
    cin>>start>>endd;
    while(scanf("%lld%lld%lld%lld",&x1,&y1,&x2,&y2)!=EOF)
        dis+=sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
 
    double ans=dis*2/1000/20;
    long long h=(long long)(ans);
    long long m=(long long)((ans-h)*60+0.5);
 
    printf("%lld:%02lld\n",h,m);
    return 0;
}

三、蓝桥杯 11届

3.1 问题描述

3.2 问题解决

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

typedef pair<double, double> PDD;
set<PDD> line; 
int res = 1;  

void compute(double x, double y)
{
    set<PDD> points; 
    PDD iter;

    for(set<PDD>::iterator l = line.begin(); l != line.end(); l ++)
    {
        double a = l->first, b = l->second;
        if(a != x) 
        {
            iter.first = (b - y) / (x - a); 
            iter.second = iter.first * x + y;
            points.insert(iter);
        } 
    }   
    res += points.size();
}

int main()
{
    int n;
    cin >> n;
    while(n --)
    {
        double x, y;
        cin >> x >> y;
        int m = line.size();
        line.insert(make_pair(x, y)); 
        if(m != line.size()) 
        {
            res ++;
            compute(x, y);
        }
    }
    cout << res << endl;
    return 0;
}