hdu2389 Rain on your Parade--HK算法 & 最大匹配数

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=2389

题意：pNum个人，uNum把伞，给定这些人和伞的二维坐标，和每个人的行进速度。问在t时间内，最多有多少人能拿到伞？

建二分图，左边是人，与右边是伞，如果这个人到伞的时间不大于t就连线，最后求最大匹配就行，不过这题用匈牙利算法会超时，所以用HK算法。

#define _CRT_SECURE_NO_DEPRECATE

#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<string>
#define INF 99999999;
using namespace std;

struct People
{
double x;
double y;
double t;
};

struct Umbrella
{
double x;
double y;
};

int T;
int time;//约束时间
int pNum;//人的数目
int uNum;//伞的数目
People people[3005];
Umbrella umbrella[3005];
int dis;
int cx[3005];
int cy[3005];
int dx[3005];
int dy[3005];
bool g[3005][3005];
bool vis[3005];

double distance(People p, Umbrella u)
{
return sqrt((p.x - u.x)*(p.x - u.x) + (p.y - u.y)*(p.y - u.y));
}

bool searchPath()
{
dis = INF;
queue<int> Q;
memset(dx, -1, sizeof(dx));
memset(dy, -1, sizeof(dy));
for (int i = 0; i < pNum; i++)
{
if (cx[i] == -1)
{
Q.push(i);
dx[i] = 0;
}
}

while (!Q.empty())
{
int u = Q.front();
Q.pop();
if (dx[u] > dis)
break;
for (int v = 0; v < uNum; v++)
{
if (g[u][v] && dy[v] == -1)
{
dy[v] = dx[u] + 1;
if (cy[v] == -1)
dis = dy[v];
else
{
dx[cy[v]] = dy[v] + 1;
Q.push(cy[v]);
}
}
}
}

return dis != INF;
}

int dfs(int u)
{
for (int v = 0; v < uNum; v++)
{
if (g[u][v] && dy[v] == dx[u] + 1 && !vis[v])
{
vis[v] = 1;
if (cy[v] != -1 && dy[v] == dis)
continue;
if (cy[v] == -1 || dfs(cy[v]))
{
cx[u] = v;
cy[v] = u;
return 1;
}
}
}
return 0;
}

int HK()
{
int ans = 0;
memset(cx, -1, sizeof(cx));
memset(cy, -1, sizeof(cy));
while (searchPath())
{
memset(vis, 0, sizeof(vis));
for (int u = 0; u < pNum; u++)
{
if (cx[u] == -1)
ans += dfs(u);
}
}
return ans;
}

int main()
{
scanf("%d", &T);
int cas = 1;
while (T--)
{
scanf("%d%d", &time, &pNum);
for (int i = 0; i < pNum; i++)
scanf("%lf%lf%lf", &people[i].x, &people[i].y, &people[i].t);
scanf("%d", &uNum);
for (int i = 0; i < uNum; i++)
scanf("%lf%lf", &umbrella[i].x, &umbrella[i].y);

memset(g, 0, sizeof(g));

for (int i = 0; i < pNum; i++)
for (int j = 0; j < uNum; j++)
if (distance(people[i], umbrella[j]) / people[i].t <= time)
g[i][j] = 1;

printf("Scenario #%d:\n%d\n\n", cas++, HK());
}
return 0;
}