I‘m Telling the Truth

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

After this year’s college-entrance exam, the teacher did a survey in his class on students’ score. There are n students in the class. The students didn’t want to tell their teacher their exact score; they only told their teacher their
rank in the province (in the form of intervals).

After asking all the students, the teacher found that some students didn’t tell the truth. For example, Student1 said he was between 5004th and 5005th, Student2 said he was between 5005th and 5006th, Student3 said he was between 5004th and 5006th, Student4
said he was between 5004th and 5006th, too. This situation is obviously impossible. So at least one told a lie. Because the teacher thinks most of his students are honest, he wants to know how many students told the truth at most.

Input

There is an integer in the first line, represents the number of cases (at most 100 cases). In the first line of every case, an integer n (n <= 60) represents the number of students. In the next n lines of every case, there are 2 numbers
in each line, X_i and Y_i (1 <= X_i <= Y_i <= 100000), means the i-th student’s rank is between X_i and Y_i, inclusive.

Output

Output 2 lines for every case. Output a single number in the first line, which means the number of students who told the truth at most. In the second line, output the students who tell the truth, separated by a space. Please note
that there are no spaces at the head or tail of each line. If there are more than one way, output the list with maximum lexicographic. (In the example above, 1 2 3;1 2 4;1 3 4;2 3 4 are all OK, and 2 3 4 with maximum lexicographic)

Sample Input

2
4
5004 5005
5005 5006
5004 5006
5004 5006
7
4 5
2 3
1 2
2 2
4 4
2 3
3 4

Sample Output

3
2 3 4
5
1 3 5 6 7

Source

2010 Asia Tianjin Regional Contest

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题目大意：每个人都说出自己所在的名次范围，问最多多少个人说的是实话，并把说实话的人的号输出出来，如果有多种情况满足的话，就输出最大的序列

ac代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int head[100100],vis[100100],cnt,ans[100100],link[100100];
int cmp(const void *a,const void *b)
{
	return *(int *)a-*(int *)b;
}
struct s
{
	int u,v,next;
}edge[100010];
void add(int u,int v)
{
	edge[cnt].u=u;
	edge[cnt].v=v;
	edge[cnt].next=head[u];
	head[u]=cnt++;
}
int dfs(int u)
{
	int i;
	for(i=head[u];i!=-1;i=edge[i].next)
	{
		int v=edge[i].v;
		if(!vis[v])
		{
			vis[v]=1;
			if(link[v]==-1||dfs(link[v]))
			{
				link[v]=u;
				return 1;
			}
		}
	}
	return 0;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		int i,j;
		cnt=0;
		memset(head,-1,sizeof(head));
		memset(link,-1,sizeof(link));
		for(i=1;i<=n;i++)
		{
			int a,b;
			scanf("%d%d",&a,&b);
			for(j=a;j<=b;j++)
				add(i,j);
		}
		int top=0;
		for(i=n;i>0;i--)
		{
			memset(vis,0,sizeof(vis));
			if(dfs(i))
			{
				ans[top++]=i;
			}
		}
		printf("%d\n",top);
		qsort(ans,top,sizeof(ans[0]),cmp);
		for(i=0;i<top;i++)
		{
			if(i)
				printf(" %d",ans[i]);
			else
				printf("%d",ans[i]);
		}
		printf("\n");
	}
}

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HDOJ题目3729 I'm Telling the Truth（二分图）