Rain on your Parade

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 655350/165535 K (Java/Others)
Total Submission(s): 4889 Accepted Submission(s): 1612

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 <= s_i <= 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

Source

HDU 2008-10 Public Contest

Recommend

lcy

题解：

就直接求二分图最大匹配，不过由于数据较大，匈牙利算法超时，所以需要用HK算法。

代码如下：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cstdlib>
  5 #include <string>
  6 #include <vector>
  7 #include <map>
  8 #include <set>
  9 #include <queue>
 10 #include <sstream>
 11 #include <algorithm>
 12 using namespace std;
 13 const int INF = 2e9;
 14 const int MOD = 1e9+7;
 15 const int MAXN = 3000+10;
 16 
 17 struct Node
 18 {
 19     int x, y, speed;
 20 }gue[MAXN], umb[MAXN];
 21 
 22 int uN, vN, t;
 23 vector<int>g[MAXN];
 24 
 25 int Mx[MAXN], My[MAXN];
 26 int dx[MAXN], dy[MAXN];
 27 int dis;
 28 bool used[MAXN];
 29 
 30 bool SearchP()
 31 {
 32     queue<int>Q;
 33     dis = INF;
 34     memset(dx, -1, sizeof(dx));
 35     memset(dy, -1, sizeof(dy));
 36     for(int i = 1; i<=uN; i++)
 37     if(Mx[i]==-1)
 38     {
 39         Q.push(i);
 40         dx[i] = 0;
 41     }
 42 
 43     while(!Q.empty())
 44     {
 45         int u = Q.front();
 46         Q.pop();
 47         if(dx[u]>dis) break;
 48         int sz = g[u].size();
 49         for(int i = 0; i<sz; i++)
 50         {
 51             int v = g[u][i];
 52             if(dy[v]==-1)
 53             {
 54                 dy[v] = dx[u] + 1;
 55                 if(My[v]==-1) dis = dy[v];
 56                 else
 57                 {
 58                     dx[My[v]] = dy[v] + 1;
 59                     Q.push(My[v]);
 60                 }
 61             }
 62         }
 63     }
 64     return dis!=INF;
 65 }
 66 
 67 bool DFS(int u)
 68 {
 69     int sz = g[u].size();
 70     for(int i = 0; i<sz; i++)
 71     {
 72         int v = g[u][i];
 73         if(!used[v] && dy[v]==dx[u]+1)
 74         {
 75             used[v] = true;
 76             if(My[v]!=-1 && dy[v]==dis) continue;
 77             if(My[v]==-1 || DFS(My[v]))
 78             {
 79                 My[v] = u;
 80                 Mx[u] = v;
 81                 return true;
 82             }
 83         }
 84     }
 85     return false;
 86 }
 87 
 88 int MaxMatch()
 89 {
 90     int res = 0;
 91     memset(Mx, -1, sizeof(Mx));
 92     memset(My, -1, sizeof(My));
 93     while(SearchP())
 94     {
 95         memset(used, false, sizeof(used));
 96         for(int i = 1; i<=uN; i++)
 97             if(Mx[i]==-1 && DFS(i))
 98                 res++;
 99     }
100     return res;
101 }
102 
103 int main()
104 {
105     int T, kase = 0;
106     scanf("%d", &T);
107     while(T--)
108     {
109         scanf("%d%d", &t, &uN);
110         for(int i = 1; i<=uN; i++)
111         {
112             scanf("%d%d%d", &gue[i].x, &gue[i].y, &gue[i].speed);
113             g[i].clear();
114         }
115 
116         scanf("%d", &vN);
117         for(int i = 1; i<=vN; i++)
118             scanf("%d%d", &umb[i].x, &umb[i].y);
119 
120         for(int i = 1; i<=uN; i++)
121         for(int j = 1; j<=vN; j++)
122         {
123             int dis = (gue[i].x-umb[j].x)*(gue[i].x-umb[j].x)
124                       +(gue[i].y-umb[j].y)*(gue[i].y-umb[j].y);
125             int s = gue[i].speed*gue[i].speed*t*t;
126             if(s>=dis) g[i].push_back(j);
127         }
128 
129         int ans = MaxMatch();
130         printf("Scenario #%d:
%d

", ++kase, ans);
131 
132     }
133 }

View Code