luogu P2526 [SHOI2001]小狗散步 二分图匹配

计算一下，对于主人经过一段路线，宠物来得及跑到哪些景点再跑回来。然后对于主人的一段路程看作一个点，与来得及的景点连边。跑一下匈牙利算法就可以了。

#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
bool flg[110];
int a[110][2],b[110][2];
int mth[110],res[110],head[110],to[11000],nxt[11000];
int cnt,n,m,ans;
bool find(int x)  
{  
    for (int i = head[x];i;i = nxt[i])  
    {  
        if (flg[to[i]] == false)  
        {  
            flg[to[i]] = true;  
            if (find(mth[to[i]]) || mth[to[i]] == 0)  
            {  
                mth[to[i]] = x;
                res[x] = to[i];  
                return true;  
            }  
        }  
    }  
    return false;  
}  
void match()
{
    for (int i = 1;i <= n - 1;i++)
    {
        memset(flg,0,sizeof(flg));
        if (find(i))
             ans++;
    }
}
void add(int x,int y)
{
    nxt[++cnt] = head[x];
    to[cnt] = y;
    head[x] = cnt;
}
int sqr(int x)
{
    return x * x;
}
double dis(int x1,int y1,int x2,int y2)
{
    return sqrt(sqr(x2 - x1) + sqr(y2 - y1));
}
int main()
{
    scanf("%d%d",&n,&m);
    ans = n;
    for (int i = 1;i <= n;i++)
        scanf("%d%d",&a[i][0],&a[i][1]);
    for (int i = 1;i <= m;i++)
        scanf("%d%d",&b[i][0],&b[i][1]);
    for (int i = 1;i <= n - 1;i++)
        for (int j = 1;j <= m;j++)
            if (2 * dis(a[i][0],a[i][1],a[i + 1][0],a[i + 1][1]) >= dis(a[i][0],a[i][1],b[j][0],b[j][1]) + dis(a[i + 1][0],a[i + 1][1],b[j][0],b[j][1]))
                add(i,j);
    match();
    printf("%d
",ans);
    for (int i = 1;i <= n;i++)
    {
        printf("%d %d ",a[i][0],a[i][1]);
        if (res[i] != 0)
            printf("%d %d ",b[res[i]][0],b[res[i]][1]); 
    }
    return 0;
}