hdu 2389 Rain on your Parade（二分图最大匹配，Hopcroft-Karp）

Rain on your Parade

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 655350/165535 K (Java/Others)

Total Submission(s): 3659    Accepted Submission(s): 1186


Problem Description

You’re giving a party in the garden of your villa by the sea. The party is a huge success, and everyone is here. It’s a warm, sunny evening, and a soothing wind sends fresh, salty air from the sea. The evening is progressing just as you had imagined. It could
be the perfect end of a beautiful day.

But nothing ever is perfect. One of your guests works in weather forecasting. He suddenly yells, “I know that breeze! It means its going to rain heavily in just a few minutes!” Your guests all wear their best dresses and really would not like to get wet, hence
they stand terrified when hearing the bad news.

You have prepared a few umbrellas which can protect a few of your guests. The umbrellas are small, and since your guests are all slightly snobbish, no guest will share an umbrella with other guests. The umbrellas are spread across your (gigantic) garden, just
like your guests. To complicate matters even more, some of your guests can’t run as fast as the others.

Can you help your guests so that as many as possible find an umbrella before it starts to pour? 

Given the positions and speeds of all your guests, the positions of the umbrellas, and the time until it starts to rain, find out how many of your guests can at most reach an umbrella. Two guests do not want to share an umbrella, however. 

 

Input

The input starts with a line containing a single integer, the number of test cases.

Each test case starts with a line containing the time t in minutes until it will start to rain (1 <=t <= 5). The next line contains the number of guests m (1 <= m <= 3000), followed by m lines containing x- and y-coordinates as well as the speed si in units
per minute (1 <= si <= 3000) of the guest as integers, separated by spaces. After the guests, a single line contains n (1 <= n <= 3000), the number of umbrellas, followed by n lines containing the integer coordinates of each umbrella, separated
by a space.

The absolute value of all coordinates is less than 10000.

 

Output

For each test case, write a line containing “Scenario #i:”, where i is the number of the test case starting at 1. Then, write a single line that contains the number of guests that can at most reach an umbrella before it starts to rain. Terminate every test
case with a blank line.

 

Sample Input

2
1
2
1 0 3
3 0 3
2
4 0
6 0
1
2
1 1 2
3 3 2
2
2 2
4 4

 

Sample Output

Scenario #1:
2

Scenario #2:
2

 
题意：有m个人，n把伞。现在给你人和伞的坐标，人的速度，距离下雨的时间，问你最多能有几个人能拿到伞。距离为直线距离

思路：此题和北大一个地鼠的题一样，不过数据量大了..匈牙利算法无限TLE

这里要用一个匈牙利算法的优化版本，叫Hopcroft-Karp

关于如何优化可以去百度，本人理解的也不是很深刻..觉得跟网络流的层次网络有点像

好像用到这个算法的题目也少之又少，所以直接当模板算了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
#define N 3002
#define INF 99999999
int m,n,t;
int nx,ny,xline
,yline
,dy
,dx
;
int vis
,dis;
struct Node
{
long long x,y,v;
}p
;
struct Edge
{
int v,next;
}edge[N*N];
int cnt,head
;
void init()
{
cnt=0;
memset(head,-1,sizeof(head));
memset(xline,-1,sizeof(xline));
memset(yline,-1,sizeof(yline));
}
void addedge(int u,int v)
{
edge[cnt].v=v;
edge[cnt].next=head[u];
head[u]=cnt++;
}
long long dist(Node a,Node b)
{
return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
int bfs()
{
queue<int>que;
dis=INF;
memset(dx,-1,sizeof(dx));
memset(dy,-1,sizeof(dy));
for(int i=1;i<=m;i++)
{
if(xline[i]==-1)
{
que.push(i);
dx[i]=0;
}
}
while(!que.empty())
{
int u=que.front();que.pop();
if(dx[u]>dis) break;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v = edge[i].v;
if(dy[v] == -1)
{
dy[v] = dx[u] + 1;
if(yline[v] == -1) dis = dy[v];
else
{
dx[yline[v]] = dy[v]+1;
que.push(yline[v]);
}
}
}
}
return dis!=INF;
}
int can(int t)
{
for(int i=head[t];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(!vis[v]&&dy[v]==dx[t]+1)
{
vis[v]=1;
if(yline[v]!=-1&&dy[v]==dis) continue;
if(yline[v]==-1||can(yline[v]))
{
yline[v]=t,xline[t]=v;
return 1;
}
}
}
return 0;
}
int Maxmatch()
{
int ans=0;
while(bfs())
{
memset(vis,0,sizeof(vis));
for(int i=1;i<=m;i++)
if(xline[i]==-1&&can(i))
ans++;
}
return ans;
}
int main()
{
int T,tot=1;
int x,y;
Node a;
scanf("%d",&T);
while(T--)
{
init();
scanf("%d %d",&t,&m);
for(int i=1;i<=m;i++)
{
scanf("%lld %lld %lld",&p[i].x,&p[i].y,&p[i].v);
p[i].v*=t;
p[i].v=p[i].v*p[i].v;
}
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%lld %lld",&a.x,&a.y);
for(int j=1;j<=m;j++)
{
long long d=dist(a,p[j]);
if(p[j].v>=d)
addedge(j,i);
}
}
int ans=Maxmatch();
printf("Scenario #%d:\n",tot++);
printf("%d\n\n",ans);
}
return 0;
}