POJ——T 1422 Air Raid
http://poj.org/problem?id=1422
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 8579 | Accepted: 5129 |
Description
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
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
Sample Input
2
4
3
3 4
1 3
2 3
3
3
1 3
1 2
2 3
Sample Output
2
1
Source
#include <cstring>
#include <cstdio> using namespace std; int n,map[][],match[];
bool vis[]; bool find(int u)
{
for(int v=;v<=n;v++)
if(map[u][v]&&!vis[v])
{
vis[v]=;
if(!match[v]||find(match[v]))
{
match[v]=u;
return true;
}
}
return false;
} int AC()
{
int t;scanf("%d",&t);
for(int m,ans;t--;)
{
scanf("%d%d",&n,&m);ans=n;
for(int u,v;m--;map[u][v]=)
scanf("%d%d",&u,&v);
for(int i=;i<=n;i++)
{
memset(vis,,sizeof(vis));
if(find(i)) ans--;
}
printf("%d\n",ans);
memset(map,,sizeof(map));
memset(match,,sizeof(match));
}
return ;
} int I_want_AC=AC();
int main(){;}
POJ——T 1422 Air Raid的更多相关文章
- POJ 1422 Air Raid(二分图匹配最小路径覆盖)
POJ 1422 Air Raid 题目链接 题意:给定一个有向图,在这个图上的某些点上放伞兵,能够使伞兵能够走到图上全部的点.且每一个点仅仅被一个伞兵走一次.问至少放多少伞兵 思路:二分图的最小路径 ...
- poj 1422 Air Raid (二分匹配)
Air Raid Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 6520 Accepted: 3877 Descript ...
- poj——1422 Air Raid
Air Raid Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 8577 Accepted: 5127 Descript ...
- poj 1422 Air Raid 最少路径覆盖
题目链接:http://poj.org/problem?id=1422 Consider a town where all the streets are one-way and each stree ...
- POJ 1422 Air Raid
题目链接: http://poj.org/problem?id=1422 Description Consider a town where all the streets are one-way a ...
- POJ 1422 Air Raid (最小路径覆盖)
题意 给定一个有向图,在这个图上的某些点上放伞兵,可以使伞兵可以走到图上所有的点.且每个点只被一个伞兵走一次.问至少放多少伞兵. 思路 裸的最小路径覆盖. °最小路径覆盖 [路径覆盖]在一个有向图G( ...
- POJ - 1422 Air Raid 二分图最大匹配
题目大意:有n个点,m条单向线段.如今问要从几个点出发才干遍历到全部的点 解题思路:二分图最大匹配,仅仅要一条匹配,就表示两个点联通,两个点联通仅仅须要选取当中一个点就可以,所以有多少条匹配.就能够减 ...
- POJ - 1422 Air Raid(DAG的最小路径覆盖数)
1.一个有向无环图(DAG),M个点,K条有向边,求DAG的最小路径覆盖数 2.DAG的最小路径覆盖数=DAG图中的节点数-相应二分图中的最大匹配数 3. /* 顶点编号从0开始的 邻接矩阵(匈牙利算 ...
- POJ Air Raid 【DAG的最小不相交路径覆盖】
传送门:http://poj.org/problem?id=1422 Air Raid Time Limit: 1000MS Memory Limit: 10000K Total Submissi ...
随机推荐
- Codeforces 667B Coat of Anticubism
链接:传送门 题意:题目balabala说了一大堆,然而并没什么卵用,给你n个数,将这个集合分割成两部分,构成三角形的两个边,让你求补充的那个边最短是多长 思路:三角形三边具有 a + b > ...
- POJ 2356 Find a multiple( 鸽巢定理简单题 )
链接:传送门 题意:题意与3370类似 注意:注意输出就ok,输出的是集合的值不是集合下标 /***************************************************** ...
- Crontab入门基础
Crontab入门基础 crontab前言 crontab是Unix和Linux用于设置周期性被执行的指令,是互联网很常用的技术,很多任务都会设置在crontab循环执行,如果不使用crontab,那 ...
- CSS3特效——六面体
<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title> ...
- Java String.replaceAll()方法
声明 以下是java.lang.String.replaceAll()方法的声明 public String replaceAll(String regex, String replacement) ...
- Objective-C 和 Core Foundation 对象相互转换
iOS同意Objective-C 和 Core Foundation 对象之间能够轻松的转换: CFStringRef aCFString = (CFStringRef)aNSString; NSSt ...
- shell文本过滤编程(一):grep和正則表達式
[版权声明:转载请保留出处:blog.csdn.net/gentleliu.Mail:shallnew at 163 dot com] Linux系统中有非常多文件,比方配置文件.日志文件.用户文件等 ...
- linux下通过命令启动多个终端运行对应的命令和程序
作者:张昌昌 在一些情况下,往往须要同一时候启动多个终端并让终端运行自己主动运行对应的命令,进而达到提高操作效率的目的.在linux下gnome-terminal启动终端命令, gnome-t ...
- JDBC连接mysql增删改查整体代码
第一种比较low:用了statment,没有用preparedstatement.另外,插入时,不灵活,不能调用参数,但是如果直接给函数形参的话就会被SQL注入攻击,所以,最好在sql语句中使用?代表 ...
- 深入分析Java中的I/O类的特征及适用场合
Java中有40多个与输入输出有关的类.假设不理清它们之间的关系.就不能灵活地运用它们. 假设从流的流向来分,可分为输入流和输出流,而输入流和输出流又都可分为字节流和字符流.因而可将Java中的I/O ...