二分图最大匹配(匈牙利算法) UVA 670 The dog task

题目传送门

/*
    题意：bob按照指定顺序行走，他的狗可以在他到达下一个点之前到一个景点并及时返回，问狗最多能走多少个景点
    匈牙利算法：按照狗能否顺利到一个景点分为两个集合，套个模板
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;

const int MAXN = 1e2 + 10;
const int INF = 0x3f3f3f3f;
struct P
{
    int x, y;
    double nxt;
}bob[MAXN], dog[MAXN];
double d[MAXN][MAXN];
bool vis[MAXN];
int lk[MAXN];
int lk2[MAXN];
vector<int> G[MAXN];

double get_dis(int x1, int y1, int x2, int y2)
{
    return sqrt ((1.0) * ((x1-x2) * (x1-x2) + (y1-y2) * (y1-y2)));
}

bool DFS(int u)
{
    for (int i=0; i<G[u].size (); ++i)
    {
        int v = G[u][i];
        if (!vis[v])
        {
            vis[v] = true;
            if (lk[v] == -1 || DFS (lk[v]))
            {
                lk[v] = u;    lk2[u] = v;    return true;
            }
        }
    }

    return false;
}

void hungary(int n)
{
    int res = 0;    memset (lk, -1, sizeof (lk));    memset (lk2, -1, sizeof (lk2));
    for (int i=1; i<n; ++i)
    {
        memset (vis, false, sizeof (vis));
        if (DFS (i))    res++;
    }

    printf ("%d
", n + res);
    for (int i=1; i<n; ++i)
    {
        printf ("%d %d ", bob[i].x, bob[i].y);
        if (lk2[i] != -1)    printf ("%d %d ", dog[lk2[i]].x, dog[lk2[i]].y);
    }
    printf ("%d %d
", bob[n].x, bob[n].y);
}

int main(void)        //UVA 670 The dog task
{
//    freopen ("UVA_670.in", "r", stdin);

    int t;    scanf ("%d", &t);
    while (t--)
    {
        int n, m;    scanf ("%d%d", &n, &m);
        for (int i=1; i<=n; ++i)
        {
            scanf ("%d%d", &bob[i].x, &bob[i].y);    G[i].clear ();
        }
        for (int i=1; i<=m; ++i)
        {
            scanf ("%d%d", &dog[i].x, &dog[i].y);
        }

        for (int i=1; i<=n; ++i)
        {
            for (int j=1; j<=m; ++j)
            {
                d[i][j] = get_dis (bob[i].x, bob[i].y, dog[j].x, dog[j].y);
            }
        }

        for (int i=1; i<n; ++i)
        {
            bob[i].nxt = get_dis (bob[i].x, bob[i].y, bob[i+1].x, bob[i+1].y);
        }

        for (int i=1; i<n; ++i)
        {
            for (int j=1; j<=m; ++j)
            {
                if (d[i][j] + d[i+1][j] <= 2 * bob[i].nxt)    G[i].push_back (j);
            }
        }

        hungary (n);
        if (t)    puts ("");
    }

    return 0;
}