HDU 1151 Air Raid（最小路径覆盖）

Air Raid

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1904    Accepted Submission(s): 1225

Problem Description

Consider a town where all the streets are one-way and each street leads from one intersection to another. It is also known that starting from an intersection and walking through town's streets you can never reach the same intersection i.e. the town's streets form no cycles.

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

 

 

Source

Asia 2002, Dhaka (Bengal)

 

 

Recommend

Ignatius.L

 

 

 

 

 

Each paratrooper lands at an intersection and can visit other intersections following the town streets.

 

 

 

是最小路径覆盖。

注意是有向边。扩展的时候建一边的边。

而且结果不要除于2

/*

HDU 1151*/



#include<stdio.h>

#include<string.h>

#include<algorithm>

#include<iostream>

using namespace std;





/* **************************************************************************

//二分图匹配（匈牙利算法的DFS实现）

//初始化：g[][]两边顶点的划分情况

//建立g[i][j]表示i->j的有向边就可以了，是左边向右边的匹配

//g没有边相连则初始化为0

//uN是匹配左边的顶点数，vN是匹配右边的顶点数

//调用：res=hungary();输出最大匹配数

//优点：适用于稠密图，DFS找增广路，实现简洁易于理解

//时间复杂度:O(VE)

//***************************************************************************/

//顶点编号从0开始的

const int MAXN=150;

int uN,vN;//u,v数目

int g[MAXN][MAXN];

int linker[MAXN];

bool used[MAXN];

bool dfs(int u)//从左边开始找增广路径

{

    int v;

    for(v=0;v<vN;v++)//这个顶点编号从0开始，若要从1开始需要修改

      if(g[u][v]&&!used[v])

      {

          used[v]=true;

          if(linker[v]==-1||dfs(linker[v]))

          {//找增广路，反向

              linker[v]=u;

              return true;

          }

      }

    return false;//这个不要忘了，经常忘记这句

}

int hungary()

{

    int res=0;

    int u;

    memset(linker,-1,sizeof(linker));

    for(u=0;u<uN;u++)

    {

        memset(used,0,sizeof(used));

        if(dfs(u)) res++;

    }

    return res;

}

//******************************************************************************/

int main()

{

    int k;

    int n;

    int u,v;

    int T;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d%d",&n,&k);

        memset(g,0,sizeof(g));

        while(k--)

        {

            scanf("%d%d",&u,&v);

            u--;

            v--;

            g[u][v]=1;

        }

        uN=vN=n;

        printf("%d\n",n-hungary());

    }

    return 0;

}

 

 

 

二分图就是建模复杂。。。。

写代码都是模板。。。。。