最小路径覆盖 hdu 1151 hdu 3335

Air Raid

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3122    Accepted Submission(s): 2027

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)

题意：有一个城镇，它的全部街道都是单行的，而且每条街道都是和两个路口相连。同一时候已知街道不会形成回路。
你的任务是编敲代码求最小数量的伞兵，这些伞兵能够訪问（visit）全部的路口。对于伞兵的起始降落点不做限制。

最小路径覆盖问题：用尽量少的不相交简单路径覆盖有向无环图的全部顶点。

最小路径覆盖数=节点数（n）- 最大匹配数（m）

#include"stdio.h"
#include"string.h"
#include"stdlib.h"
#include"algorithm"
using namespace std;
#define N 150
int g[N][N];
int mark[N],link[N],n;
int find(int k)
{
    int i;
    for(i=1;i<=n;i++)
    {
        if(g[k][i]&&!mark[i])
        {
            mark[i]=1;
            if(link[i]==-1||find(link[i]))
            {
                link[i]=k;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int T,i,m,u,v;
    scanf("%d",&T);
    while(T--)
    {
        memset(link,-1,sizeof(link));
        memset(g,0,sizeof(g));
        scanf("%d%d",&n,&m);
        while(m--)
        {
            scanf("%d%d",&u,&v);
            g[u][v]=1;
        }
        int ans=0;
        for(i=1;i<=n;i++)
        {
            memset(mark,0,sizeof(mark));
            ans+=find(i);
        }
        printf("%d
",n-ans);
    }
    return 0;
}

Divisibility

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1412    Accepted Submission(s): 516

Problem Description

As we know,the fzu AekdyCoin is famous of math,especially in the field of number theory.So,many people call him "the descendant of Chen Jingrun",which brings him a good reputation.
AekdyCoin also plays an important role in the ACM_DIY group,many people always ask him questions about number theory.One day,all members urged him to conduct a lesson in the group.The rookie daizhenyang is extremely weak at math,so he is delighted.
However,when AekdyCoin tells us "As we know, some numbers have interesting property. For example, any even number has the property that could be divided by 2.",daizhenyang got confused,for he don't have the concept of divisibility.He asks other people for help,first,he randomizely writes some positive integer numbers,then you have to pick some numbers from the group,the only constraint is that if you choose number a,you can't choose a number divides a or a number divided by a.(to illustrate the concept of divisibility),and you have to choose as many numbers as you can.
Poor daizhenyang does well in neither math nor programming.The responsibility comes to you!

 

Input

An integer t,indicating the number of testcases,
For every case, first a number n indicating daizhenyang has writen n numbers(n<=1000),then n numbers,all in the range of (1...2^63-1).

 

Output

The most number you can choose.

 

Sample Input

1
3
1 2 3 

 

Sample Output

2


Hint:
If we choose 2 and 3,one is not divisible by the other,which is the most number you can choose.

 

#include"stdio.h"
#include"string.h"
#include"stdlib.h"
#define N 1005
int g[N][N],n;
int mark[N],link[N];
int find(int k)       //匈牙利算法求最大匹配数目
{
    int i;
    for(i=0;i<n;i++)
    {
        if(g[k][i]&&!mark[i])
        {
            mark[i]=1;
            if(link[i]==-1||find(link[i]))
            {
                link[i]=k;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,j,T;
    __int64 a[N];
    scanf("%d",&T);
    while(T--)
    {
        memset(link,-1,sizeof(link));
        memset(g,0,sizeof(g));
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%I64d",&a[i]);
            for(j=0;j<i;j++)
            {
                if(a[i]%a[j]==0||a[j]%a[i]==0)
                {
                    g[i][j]=1;
                }
            }
        }
        int ans=0;
        for(i=0;i<n;i++)
        {
            memset(mark,0,sizeof(mark));
            ans+=find(i);
        }
        printf("%d
",n-ans);
    }
    return 0;
}