hdu 1151 Air Raid - 二分匹配
With these assumptions your task is to write a program that finds the minimum number of paratroopers that can descend on the town and visit all the intersections of this town in such a way that more than one paratrooper visits no intersection. Each paratrooper lands at an intersection and can visit other intersections following the town streets. There are no restrictions about the starting intersection for each paratrooper.
Input
Your program should read sets of data. The first line of the input file contains the number of the data sets. Each data set specifies the structure of a town and has the format:
no_of_intersections
no_of_streets
S1 E1
S2 E2
......
Sno_of_streets Eno_of_streets
The first line of each data set contains a positive integer no_of_intersections (greater than 0 and less or equal to 120), which is the number of intersections in the town. The second line contains a positive integer no_of_streets, which is the number of streets in the town. The next no_of_streets lines, one for each street in the town, are randomly ordered and represent the town's streets. The line corresponding to street k (k <= no_of_streets) consists of two positive integers, separated by one blank: Sk (1 <= Sk <= no_of_intersections) - the number of the intersection that is the start of the street, and Ek (1 <= Ek <= no_of_intersections) - the number of the intersection that is the end of the street. Intersections are represented by integers from 1 to no_of_intersections.
There are no blank lines between consecutive sets of data. Input data are correct.
Output
The result of the program is on standard output. For each input data set the program prints on a single line, starting from the beginning of the line, one integer: the minimum number of paratroopers required to visit all the intersections in the town.
Sample Input
2
4
3
3 4
1 3
2 3
3
3
1 3
1 2
2 3
Sample Output
2
1
最小路径覆盖裸模型。答案 = 点数 - 最大匹配数。
Code
/**
* hdu
* Problem#1151
* Accepted
* Time:0ms
* Memory:1680k
*/
#include<iostream>
#include<cstdio>
#include<ctime>
#include<cctype>
#include<cstring>
#include<cstdlib>
#include<fstream>
#include<sstream>
#include<algorithm>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<vector>
#include<stack>
#ifndef WIN32
#define Auto "%lld"
#else
#define Auto "%I64d"
#endif
using namespace std;
typedef bool boolean;
const signed int inf = (signed)((1u << ) - );
const double eps = 1e-;
#define smin(a, b) a = min(a, b)
#define smax(a, b) a = max(a, b)
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
template<typename T>
inline boolean readInteger(T& u){
char x;
int aFlag = ;
while(!isdigit((x = getchar())) && x != '-' && x != -);
if(x == -) {
ungetc(x, stdin);
return false;
}
if(x == '-'){
x = getchar();
aFlag = -;
}
for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
ungetc(x, stdin);
u *= aFlag;
return true;
} ///map template starts
typedef class Edge{
public:
int end;
int next;
Edge(const int end = , const int next = ):end(end), next(next){}
}Edge; typedef class MapManager{
public:
int ce;
int *h;
Edge *edge;
MapManager(){}
MapManager(int points, int limit):ce() {
h = new int[(const int)(points + )];
edge = new Edge[(const int)(limit + )];
memset(h, , sizeof(int) * (points + ));
}
inline void addEdge(int from, int end){
edge[++ce] = Edge(end, h[from]);
h[from] = ce;
}
inline void addDoubleEdge(int from, int end){
addEdge(from, end);
addEdge(end, from);
}
Edge& operator [] (int pos) {
return edge[pos];
}
inline void clear() {
delete[] h;
delete[] edge;
}
}MapManager;
#define m_begin(g, i) (g).h[(i)]
///map template ends int T;
int n, m;
MapManager g; inline void init() {
readInteger(n);
readInteger(m);
g = MapManager(n, m);
for(int i = , a, b; i < m; i++) {
readInteger(a);
readInteger(b);
g.addEdge(a, b);
}
} int* match;
boolean* vis;
boolean dfs(int node) {
for(int i = m_begin(g, node); i; i = g[i].next) {
int &e = g[i].end;
if(vis[e]) continue;
vis[e] = true;
if(match[e] == - || dfs(match[e])) {
match[e] = node;
return true;
}
}
return false;
} int res;
inline void solve() {
res = n;
vis = new boolean[ * n + ];
match = new int[ * n + ];
memset(match, -, sizeof(int) * ( * n + ));
for(int i = ; i <= n; i++) {
memset(vis, false, sizeof(boolean) * ( * n + ));
if(dfs(i)) res--;
}
printf("%d\n", res);
delete[] vis;
delete[] match;
} int main() {
readInteger(T);
while(T--) {
init();
solve();
}
return ;
}
hdu 1151 Air Raid - 二分匹配的更多相关文章
- hdu 1151 Air Raid(二分图最小路径覆盖)
http://acm.hdu.edu.cn/showproblem.php?pid=1151 Air Raid Time Limit: 1000MS Memory Limit: 10000K To ...
- hdu 1151 Air Raid DAG最小边覆盖 最大二分匹配
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1151 题目大意: 城镇之间互相有边,但都是单向的,并且不会构成环,现在派伞兵降落去遍历城镇,问最少最少 ...
- hdu 1150 Machine Schedule hdu 1151 Air Raid 匈牙利模版
//两道大水……哦不 两道结论题 结论:二部图的最小覆盖数=二部图的最大匹配数 有向图的最小覆盖数=节点数-二部图的最大匹配数 //hdu 1150 #include<cstdio> #i ...
- poj 1422 Air Raid (二分匹配)
Air Raid Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 6520 Accepted: 3877 Descript ...
- hdu1151 Air Raid 二分匹配
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1151 求最小路径覆盖 二分图最小路径覆盖=点的个数-最大匹配. 代码: #include<ios ...
- hdu - 1151 Air Raid(有向无环图的最小路径覆盖)
http://acm.hdu.edu.cn/showproblem.php?pid=1151 在一个城市里有n个地点和k条道路,道路都是单向的,并且不存在环.(DAG) 现在伞兵需要去n个地点视察,伞 ...
- HDU 1151 Air Raid(最小路径覆盖)
题目大意: 有n个城市,m条道路,城市的道路是单向. 现在我们的伞兵要降落在城市里,然后我门的伞兵要搜索所有道路.问我们最少占领多少个城市就可以搜索所有的道路了. 我们可以沿着道路向前走到达另一个城 ...
- (step6.3.4)hdu 1151(Air Raid——最小路径覆盖)
题意: 一个镇里所有的路都是单向路且不会组成回路. 派一些伞兵去那个镇里,要到达所有的路口,有一些或者没有伞兵可以不去那些路口,只要其他人能完成这个任务.每个在一个路口着陆了的伞兵可以沿着街去 ...
- HDU 1151 - Air Raid
很明显求最小路径覆盖 就是求最大匹配 #include <iostream> #include <cstdio> #include <cstring> #inclu ...
随机推荐
- VS 星期作业 if else的应用 做一个受不受异性欢迎的小程序
static void Main(string[] args) { //漏掉代码 输入错误 进行提示! string T1, T2, T3, T4, T5, T6, T7, T8, T9, T10=& ...
- vmvare 将主机的文件复制到虚拟机系统中 安装WMware tools
在虚拟机里的ubuntu这里找到VMware tools包
- STL之List容器
1.List容器 1) list是一个双向链表容器,可高效地进行插入删除元素. 2)list不可以随机存取元素,所以不支持at.(pos)函数与[]操作符.It++(ok) it+5(err) 3)头 ...
- 多线程实现Thread.Start()与ThreadPool.QueueUserWorkItem两种方式对比
Thread.Start(),ThreadPool.QueueUserWorkItem都是在实现多线程并行编程时常用的方法.两种方式有何异同点,而又该如何取舍? 写一个Demo,分别用两种方式实现.观 ...
- CSS选择符-----属性选择符
Element[att] 选择具有att属性的E元素 <!DOCTYPE html> <html> <head> <meta charset=" ...
- 从零开始学习MVC
其实在学校时,已经开设了MVC这门课程,教材由授课老师自己编纂,是和微软的音乐商店相似的一个书店项目,当时无法理解 Linq.Lambda , 只记得是按照老师的方法,复制+粘贴,不明其意,亦不知其理 ...
- 20165305 苏振龙《Java程序设计》第七周学习总结
第十一章 JDBC技术在数据库开发中占有很重要的地位,JDBC操作不同的数据库仅仅是连接方式上的差异而已,使用JDBC的应用程序一旦和数据库建立连接,就可以使用JDBC提供的API操作数据库. 当查询 ...
- Vector集合——单列集合的“祖宗”类
是实现可增长的对象数组:所以底层也是数组: 与collection集合不同的是,vector是同步的,意味着是单线程的,意味着效率低,速度慢, 所以在jdk1.2版本之后被ArrayList集合所取代 ...
- NGINX的几个应用场景
NGINX的几个应用场景 两个参考地址: NGINX的百度百科:https://baike.baidu.com/item/nginx/3817705?fr=aladdin NGINX的中文网站:htt ...
- tensorflow学习5----GAN模型初探
生成模型: 通过观测学习样本和标签的联合概率分布P(X,Y)进行训练,训练好的模型能够生成符合样本分布的新数据,在无监督学习方面,生成式模型能够捕获数据的高阶相关性,通过学习真实数据的本质特征,刻画样 ...