K - Treasure Exploration （最大路径覆盖）

最新推荐文章于 2025-01-08 16:56:26 发布

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#路径覆盖 #二分匹配

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一家名为EUC的公司计划使用最少数量的机器人探索火星上的未知区域，该区域被认为充满了宝藏。通过使用路径覆盖算法来确保所有地点都能被探索到，并通过编程技巧优化机器人的路径规划。

K - Treasure Exploration

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

Sample Output

1
1
2

思路：

由题 “explore all the points” 得用路径覆盖做。

与最小路径覆盖不同，这题是最大路径覆盖。因为题中说 “You should notice that the roads of two different robots may contain some same point.”——（不同机器人的道路可能包含一些相同点），所以用 floyd 把间接相邻的点连接起来。

代码：

#include<map>
#include<queue>
#include<math.h>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define MM 507
int line[MM][MM],book[MM],man[MM];
int n,m;
bool Find(int x)
{
    for(int i=1; i<=n; i++)
    {
        if(line[x][i] && !book[i])
        {
            book[i]=1;
            if(!man[i] || Find(man[i]))
            {
                man[i]=x;
                return true;
            }
        }
    }
    return false;
}
int match()
{
    int ans=0;
    for(int i=1; i<=n; i++)
    {
        mem(book,0);
        if(Find(i))
            ans++;
    }
    return ans;
}
int main()
{
    while(~scanf("%d%d",&n,&m)&&(n||m))
    {
        mem(line,0);
        mem(man,0);
        int x,y;
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d",&x,&y);
            line[x][y]=1;
        }
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                   if(!line[i][j])
                   for(int k=1;k<=n;k++)
                   if(line[i][k]&&line[k][j])
                   line[i][j]=1;
        printf("%d\n",n-match());
    }
    return 0;
}