hdu1285+hdu2467(拓扑排序)

Posted 2020-06-14

确定比赛名次

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10604    Accepted Submission(s): 4150

Problem Description

有N个比赛队（1<=N<=500）。编号依次为1。2，3。。。

。。

，N进行比赛，比赛结束后。裁判委员会要将全部參赛队伍从前往后依次排名，但如今裁判委员会不能直接获得每一个队的比赛成绩。仅仅知道每场比赛的结果，即P1赢P2，用P1。P2表示，排名时P1在P2之前。如今请你编程序确定排名。

 

Input

输入有若干组，每组中的第一行为二个数N（1<=N<=500），M；当中N表示队伍的个数，M表示接着有M行的输入数据。接下来的M行数据中，每行也有两个整数P1。P2表示即P1队赢了P2队。

 

Output

给出一个符合要求的排名。

输出时队伍号之间有空格，最后一名后面没有空格。

其它说明：符合条件的排名可能不是唯一的。此时要求输出时编号小的队伍在前；输入数据保证是正确的，即输入数据确保一定能有一个符合要求的排名。

 

Sample Input

4 3
1 2
2 3
4 3

 

Sample Output

1 2 4 3

 

Author

SmallBeer(CML)

 

Source

杭电ACM集训队训练赛（VII）

 

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题解：
      本题是道比較裸的拓扑排序题。注意重边即可了。还有本题符合条件的排名可能不是唯一的，此时要求输出时编号小的队伍在。其次就是输出格式的限定！

#include<iostream>
#include<cstring>
using namespace std;

const int MAXN=500+10;
int inDev[MAXN];
bool visited[MAXN];
bool g[MAXN][MAXN];
int n,m;

void init()
{
	memset(g,0,sizeof(g));
	memset(visited,0,sizeof(visited));
	memset(inDev,0,sizeof(inDev));
}

void input()
{
	int a,b;
	for(int i=0;i<m;i++)
	{
		scanf("%d%d",&a,&b);
		if(!g[a][b])
		{
			g[a][b]=true;
			++inDev[b];
		}
	}
}

void topSort()  
{  
    int i,tag=0;  
    while(tag<n)  
    {  
		for(i=1;i<=n;i++)
			if(!visited[i]&&0==inDev[i])
				break;
		if(tag)
			printf(" ");
		++tag;
		printf("%d",i);
		visited[i]=true;
		for(int j=1;j<=n;j++)
			if(g[i][j])
				--inDev[j];
    }  
	printf("\n");
}  

int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		init();
		input();
		topSort();
	}
	return 0;
}

Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3580    Accepted Submission(s): 1085

Problem Description

Dandelion‘s uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewards.
The workers will compare their rewards ,and some one may have demands of the distributing of rewards ,just like a‘s reward should more than b‘s.Dandelion‘s unclue wants to fulfill all the demands, of course ,he wants to use the least money.Every work‘s reward will be at least 888 , because it‘s a lucky number.

 

Input

One line with two integers n and m ,stands for the number of works and the number of demands .(n<=10000,m<=20000)
then m lines ,each line contains two integers a and b ,stands for a‘s reward should be more than b‘s.

 

Output

For every case ,print the least money dandelion ‘s uncle needs to distribute .If it‘s impossible to fulfill all the works‘ demands ,print -1.

 

Sample Input

2 1
1 2
2 2
1 2
2 1

 

Sample Output

1777
-1

 

Author

dandelion

 

Source

曾是惊鸿照影来

 

题解：
      本题同上也是道比較easy看出的拓扑排序题，不同的建立拓扑结构图是应该逆向见图。这样可以保证前者比后者大。

#include<iostream>
#include<vector>
#include<cstring>
#include<queue>
using namespace std;

const int MAXN=10000+10;
vector<int>g[MAXN];
int inDev[MAXN];
int money[MAXN];
int n,m;

void init()
{
	for(int i=0;i<MAXN;i++)
	{
		g[i].clear();
		money[i]=888;
	}
	memset(inDev,0,sizeof(inDev));
}

void input()
{
	int a,b;
	for(int i=0;i<m;i++)
	{
		scanf("%d%d",&a,&b);
		g[b].push_back(a);
		++inDev[a];
	}
}

void topSort()
{
	queue<int>q;
	int tag=0,ans=0;
	for(int i=1;i<=n;i++)
	{
		if(inDev[i]==0)		
			q.push(i);
	}
	while(!q.empty())
	{
		++tag;
		int now=q.front();
		ans+=money[now];
		q.pop();
		for(int i=0;i<g[now].size();i++)
		{
			int next=g[now][i];
			if(--inDev[next]==0)
			{
				q.push(next);				
				money[next]=money[now]+1;
			}
		}
	}
	if(tag!=n)
		ans=-1;
	printf("%d\n",ans);
}

int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		init();
		input();
		topSort();
	}
	return 0;
}

 

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