poj 2594 Treasure Exploration

 

 

 

 

就是可以多次经过，需要用floyed缩点，与1422有所不同,是为了让每个点都只经历过一次！

网上大牛的解释

在一个ＰＸＰ的有向图中，路径覆盖就是在图中找一些路经，使之覆盖了图中的所有顶点，且任何一个顶点有且只有一条路径与之关联；（如果把这些路径中的每条路径从它的起始点走到它的终点，那么恰好可以经过图中的每个顶点一次且仅一次）；如果不考虑图中存在回路，那么每每条路径就是一个弱连通子集．

由上面可以得出：

１.一个单独的顶点是一条路径；

　 ２．如果存在一路径p1,p2,......pk，其中p1 为起点，pk为终点，那么在覆盖图中，顶点p1,p2,......pk不再与其它的顶点之间存在有向边．

　 最小路径覆盖就是找出最小的路径条数，使之成为Ｐ的一个路径覆盖．

路径覆盖与二分图匹配的关系（必须是没有圈的有向图）：

最小路径覆盖＝｜Ｐ｜－最大匹配数； 

当十字交叉的时候，例如1->2, 2->3,4->2,2->5;
如果直接用最小路径覆盖的话，先找出路径1->2->3,然后2就被删掉了，从而最终结果变为3。但是这题很明显是2.

注意！！！这道题目之所以用传递闭包，是因为每个点可以经过多次！！！

#include<cstdio>
#include<cstring>
const int N=510;
int n,m;
int map[N][N];
int link[N];
bool used[N];
void floyed()
{
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        if(i!=k)
        {
            for(int j=1;j<=n;j++)
            if(i!=j&&k!=j)
            {
                if(map[i][k]&&map[k][j])   map[i][j]=true;
            }
        }
    }
}
bool can(int t)
{
    for(int i=1;i<=n;i++)
    {
        if(used[i]==0&&map[t][i])
        {
            used[i]=1;
            if(link[i]==-1||can(link[i]))
            {
                link[i]=t;
                return true;
            }
        }
    }
    return false;
}
int MaxMatch()
{
    int sum=0;
    memset(link,-1,sizeof(link));
    for(int i=1;i<=n;i++)
    {
        memset(used,0,sizeof(used));
        if(can(i))   sum++;
    }
    return sum;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)  break;
        memset(map,0,sizeof(map));
        for(int i=0;i<m;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            map[x][y]=true;
        }
        floyed();
        printf("%d/n",n-MaxMatch());
    }
    return 0;
}

Treasure Exploration

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 3735		Accepted: 1383

Description

Have you ever read any book about treasure exploration? Have you ever see any film about treasure exploration? Have you ever explored treasure? If you never have such experiences, you would never know what fun treasure exploring brings to you.  
Recently, a company named EUC (Exploring the Unknown Company) plan to explore an unknown place on Mars, which is considered full of treasure. For fast development of technology and bad environment for human beings, EUC sends some robots to explore the treasure.  
To make it easy, we use a graph, which is formed by N points (these N points are numbered from 1 to N), to represent the places to be explored. And some points are connected by one-way road, which means that, through the road, a robot can only move from one end to the other end, but cannot move back. For some unknown reasons, there is no circle in this graph. The robots can be sent to any point from Earth by rockets. After landing, the robot can visit some points through the roads, and it can choose some points, which are on its roads, to explore. You should notice that the roads of two different robots may contain some same point.  
For financial reason, EUC wants to use minimal number of robots to explore all the points on Mars.  
As an ICPCer, who has excellent programming skill, can your help EUC?

Input

The input will consist of several test cases. For each test case, two integers N (1  <= N  <= 500) and M (0  <= M  <= 5000) are given in the first line, indicating the number of points and the number of one-way roads in the graph respectively. Each of the following M lines contains two different integers A and B, indicating there is a one-way from A to B (0  < A, B  <= N). The input is terminated by a single line with two zeros.

Output

For each test of the input, print a line containing the least robots needed.

Sample Input

1 0
2 1
1 2
2 0
0 0

Sample Output

1
1
2

Source

POJ Monthly--2005.08.28,Li Haoyuan