P2526 [SHOI2001]小狗散步 二分图+平面坐标系建图

P2526 [SHOI2001]小狗散步

题意：给出n个点，是人和小狗的共同路线，再给出m个点，是小狗喜欢的景点。小狗速度是人的两倍，可能会在中途去看景点，问最多看多少的景点。

做法：由于数据范围很小，直接暴力将每一个位置可以到达的景点和该位置连接起来。连完之后一个最大匹配即可
路径还原的时候需要注意要用inv而不是linker

一开始还WA了一发，因为最开始的时候maxn只开了100。
AC代码如下：

/**
 *  Author1: low-equipped w_udixixi
 *  Author2: Sher丶lock
 *  Date :2020-01-15
 **/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<queue>
#include<map>
#define ll long long
#define pb push_back
#define rep(x,a,b) for (int x=a;x<=b;x++)
#define repp(x,a,b) for (int x=a;x<b;x++)
#define W(x) printf("%d\n",x)
#define WW(x) printf("%lld\n",x)
#define pi 3.14159265358979323846
#define mem(a,x) memset(a,x,sizeof a)
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
using namespace std;
const int maxn=1000+7;
const int INF=1e9;
const ll INFF=1e18;
struct node
{
    int x,y;
}p[maxn];
int mapp[maxn][maxn],n,m,inv[maxn];
bool vis[maxn];
int linker[maxn];
bool dfs(int u)
{
    for (int i=n;i<=n+m;i++)
    {
        if (mapp[u][i]&&!vis[i])
        {
            vis[i]=true;
            if (linker[i]==-1||dfs(linker[i]))
            {
                linker[i]=u;
              	inv[u]=i;
                return true;
            }
        }
    }
    return false;
}
int count_()
{
    int ans=0;
    for (int i=1;i<n;i++)
    {
        mem(vis,false);
        if (dfs(i))ans++;
    }
    return ans;
}
double dis(node a,node b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}
int main()
{
    mem(linker,-1);
    mem(mapp,0);
    scanf("%d%d",&n,&m);
    rep(i,1,n+m)scanf("%d%d",&p[i].x,&p[i].y);
    rep(i,1,n-1)
    {
        double D=dis(p[i],p[i+1]);
        rep(j,n+1,n+m)
        {
            double d1=dis(p[i],p[j]);
            double d2=dis(p[i+1],p[j]);
            if (d1+d2<=2*D)mapp[i][j]=1;
        }
    }
    int x=count_();
    cout<<n+x<<endl;
    rep(i,1,n)
    {
        if (inv[i]!=0)cout<<p[i].x<<" "<<p[i].y<<" "<<p[inv[i]].x<<" "<<p[inv[i]].y<<" ";
        else cout<<p[i].x<<" "<<p[i].y<<" ";
    }
    cout<<endl;
}