Treasure Exploration

最新推荐文章于 2020-03-02 17:04:41 发布

原创最新推荐文章于 2020-03-02 17:04:41 发布 · 233 阅读

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本文介绍了一种用于最小化机器人数量的算法，旨在探索火星未知区域的每个角落。通过使用图论和匹配算法，该方法能够确定最少的机器人需求，以确保所有地点都能被覆盖。文章详细解释了算法流程，包括输入输出格式，以及一个示例输入输出。

题目描述：

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?

输入：

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.

输出：

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

样例输入：

样例输出：

1
1
2

code：

有是一个网络流，发现纯套模板果然是不行的

然后错了十几次吧

总算是对了

目前还在理解，有机会更新

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
int n,m;
const int maxn=650;
int match[maxn];
bool visit[maxn];
int a[maxn][maxn];
bool dfs(int v){
	for(int i=1;i<=n;i++){
		if(a[v][i]==1&&visit[i]==0){
			visit[i]=1;
			if(match[i]<0||dfs(match[i])){
				match[i]=v;
				return true;
			}
		}
	}
	return false;
}
int findd(){
	int res=0;
	memset(match,-1,sizeof(match));
	for(int v=1;v<=n;v++){
		memset(visit,0,sizeof(visit));
		if(dfs(v))res++;
	}
	return res;
}
int main()
{
	while(~scanf("%d%d",&n,&m)){
		if(n==0&&m==0)break;
		int u,v;
		memset(a,0,sizeof(a));
		while(m--){
			scanf("%d%d",&u,&v);
			a[u][v]=1;
		}
		for(int k=1;k<=n;k++){
			for(int i=1;i<=n;i++){
				for(int j=1;j<=n;j++){
					if(a[i][k]==1&&a[k][j]==1)
						a[i][j]=1;
				}
			}
		}
		printf("%d\n",n-findd());
	}
	return 0;
 }