HDU2389(二分图匹配Hopcroft-Carp算法)
2015-10-26 19:31
357 查看
Rain on your Parade
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 655350/165535 K (Java/Others)
Total Submission(s): 3310 Accepted Submission(s): 1066
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 Source HDU 2008-10 Public Contest 这道题用匈牙利算法会超时,匈牙利算法复杂度O(V*E) Hopcroft-Carp算法复杂度O(sqrt(V)*E)
/* ID: LinKArftc PROG: 2389.cpp LANG: C++ */ #include <map> #include <set> #include <cmath> #include <stack> #include <queue> #include <vector> #include <cstdio> #include <string> #include <utility> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define eps 1e-8 #define randin srand((unsigned int)time(NULL)) #define input freopen("input.txt","r",stdin) #define debug(s) cout << "s = " << s << endl; #define outstars cout << "*************" << endl; const double PI = acos(-1.0); const double e = exp(1.0); const int inf = 0x3f3f3f3f; const int INF = 0x7fffffff; typedef long long ll; const int maxn = 3010; struct Node { double x, y, speed; Node() {} Node(double _x, double _y) : x(_x), y(_y) {} Node(double _x, double _y, double _s) : x(_x), y(_y), speed(_s) {} } men[maxn], un[maxn]; vector<int>G[maxn]; int uN, vN; int Mx[maxn],My[maxn]; int dx[maxn],dy[maxn]; int dis; bool used[maxn]; bool SearchP() { queue<int>Q; dis = INF; memset(dx,-1,sizeof(dx)); memset(dy,-1,sizeof(dy)); for(int i = 1 ; i <= uN; i++) if(Mx[i] == -1) { Q.push(i); dx[i] = 0; } while(!Q.empty()) { int u = Q.front(); Q.pop(); if(dx[u] > dis)break; int sz = G[u].size(); for(int i = 0;i < sz;i++) { int v = G[u][i]; if(dy[v] == -1) { dy[v] = dx[u] + 1; if(My[v] == -1)dis = dy[v]; else { dx[My[v]] = dy[v] + 1; Q.push(My[v]); } } } } return dis != INF; } bool DFS(int u) { int sz = G[u].size(); for(int i = 0;i < sz;i++) { int v = G[u][i]; if(!used[v] && dy[v] == dx[u] + 1) { used[v] = true; if(My[v] != -1 && dy[v] == dis)continue; if(My[v] == -1 || DFS(My[v])) { My[v] = u; Mx[u] = v; return true; } } } return false; } int MaxMatch() { int res = 0; memset(Mx,-1,sizeof(Mx)); memset(My,-1,sizeof(My)); while(SearchP()) { memset(used,false,sizeof(used)); for(int i = 1;i <= uN;i++) if(Mx[i] == -1 && DFS(i)) res++; } return res; } int main() { //input; int T, t, _t = 1; scanf("%d", &T); while (T --) { scanf("%d", &t); scanf("%d", &uN); for (int i = 1; i <= uN; i ++) scanf("%lf %lf %lf", &men[i].x, &men[i].y, &men[i].speed); scanf("%d", &vN); for (int i = 1; i <= vN; i ++) scanf("%lf %lf", &un[i].x, &un[i].y); for (int i = 1; i <= uN; i ++) { G[i].clear(); for (int j = 1; j <= vN; j ++) { if (sqrt(fabs(men[i].x - un[j].x) * fabs(men[i].x - un[j].x) + fabs(men[i].y - un[j].y) * fabs(men[i].y - un[j].y)) - men[i].speed * t < eps) G[i].push_back(j); } } printf("Scenario #%d:\n%d\n\n", _t ++, MaxMatch()); } return 0; }
相关文章推荐
- centos6.6中mysql5.6主从复制
- Virtual 下安装CentOs6.4
- 记使用linux命令行编译c语言步骤
- shell条件判断小知识
- POJ 3050 Hopscotch
- linux——如何在linux下让系统定时自动重启(关机)
- 快速解决Android中的selinux权限问题
- linux——在linux下查看U盘
- opencv矩阵cvDet,cvDit,cvDotProduct,cvEigenVV and cvFlip
- docker简单教程
- windows下使用 Secure Shell Client工具操作linux常用命令
- linux命令expect实现ssh登陆
- hadoop reducer不执行
- windows下使用 Secure Shell Client工具操作linux常用命令
- 10 个给 Linux 用户的有用工具
- Linux系统的内存分配
- Linux进程间通信——使用共享内存
- 【linux】系统初始化的shell脚本
- ATM-PROGRAM 关于Proprties的问题
- 从此使用linux系统,但是QQ是必不可少的!!该篇文章方法成功!!!已验证!!!!!