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]1006F Xor Path
双向搜索. 水div3的时候最后一道题由于C题死活看不懂题 来不及做F了Orz.. 因为n,m是20,双向搜索一下,求个到中间的Xor值的方案,统计一下即可. 时间复杂度\(O(2^{21})\) 好 ...
- nginx 过滤zip 类型的文件
http://www.cnblogs.com/bass6/p/5500660.html
- python 多列表对应的位置的值形成一个新的列表
list1 = [1, 2, 3, 4, 5] list2 = ['a','b', 'c', 'd', 'e'] list3 = [1, 2, 3, 4, 5] multi_list = map(li ...
- LaTeX 图片色偏解决方法
本系列文章由 @YhL_Leo 出品,转载请注明出处. 文章链接: http://blog.csdn.net/yhl_leo/article/details/50327113 在LaTeX的编辑模式中 ...
- akka 原理分析优秀博客
http://www.nyankosama.com/2014/12/15/akka-source/ http://blog.csdn.net/aigoogle/article/details/4210 ...
- 【linux】——centos 分辨率配置
用过centos的朋友肯定知道centos在默认安装的时候显示器的分辨率只有800*600,但是我们想把改成1024*768或者更大,怎么办呢,我也是试过了才知道,首先打开系统-管理-显示-硬件-显示 ...
- (一)Eureka 服务的注册与发现
(一)服务的注册于发现(eureka); Eureka Server: 服务注册中心,负责服务列表的注册.维护和查询等功能 在Idea里,新建项目,选择Spring initializer. 下面的p ...
- 【学习】java下实现调用oracle的存储过程和函数
在oracle下创建一个test的账户,然后按一下步骤执行: 1.创建表:STOCK_PRICES --创建表格CREATETABLE STOCK_PRICES( RIC VARCHAR(6) PRI ...
- Dbf文件操作
package cn.com.szhtkj.util; import java.io.File; import java.io.IOException; import java.lang.reflec ...
- 创建私有CA, 加密解密基础, PKI, SSL
发现了一篇图文并茂的超棒的 加密解密, CA, PKI, SSL 的基础文章, 链接如下:https://blog.csdn.net/lifetragedy/article/details/5223 ...