【HDU】3729 I'm Telling the Truth 离散+最大流

传送门：【HDU】3729 I'm Telling the Truth

题目分析：我看这么大的数据范围，如果普通二分肯定要超时的啊。。。然后就敲了一个离散化+最大流了。。。

但是我网上看他们的题解，都是裸裸的开一个100万的数组啊！！！还比我离散的网络流还快啊啊啊！！于是我就测一次给的区间有多大（如果超出一定范围就拿一个变量除以0让报RE），第一次10000没事，然后1000。。还是没事。。。然后100仍旧没事。。我都怀疑是不是不报RE了。。。。火了，5！还是没事。。。。不管了直接1，终于给我报RE了！！！！！！最后的测验结果是：每次给的区间范围不超过5！！！！！

我不多说。。数据真是弱。。。

网络流就离散化区间就好了，很简单的。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REPV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define clear( a , x ) memset ( a , x , sizeof a )
#define copy( a , x ) memcpy ( a , x , sizeof a )

const int MAXN = 60 + 120 + 2 ;
const int MAXE = 14400 + 5 ;
const int MAXQ = 200 ;
const int INF = 0x3f3f3f3f ;

struct Edge {
	int v , c , n ;
	Edge () {}
	Edge ( int var , int cost , int next ) :
		v ( var ) , c ( cost ) , n ( next ) {}
} ;

struct Node {
	int l , r ;
	void input () {
		scanf ( "%d%d" , &l , &r ) ;
	}
} ;

struct NetWork {
	Edge E[MAXE] ;
	int H[MAXN] , cntE ;
	int d[MAXN] , num[MAXN] , cur[MAXN] , pre[MAXN] ;
	int Q[MAXQ] , head , tail ;
	int s , t , nv ;
	int n , m ;
	int flow ;
	
	//-------------
	int a[MAXN] ;
	Node A[MAXN] ;
	//-------------
	
	void init () {
		cntE = 0 ;
		clear ( H , -1 ) ;
	}
	
	void addedge ( int u , int v , int c ) {
		E[cntE] = Edge ( v , c , H[u] ) ;
		H[u] = cntE ++ ;
		E[cntE] = Edge ( u , 0 , H[v] ) ;
		H[v] = cntE ++ ;
	}
	
	void rev_bfs () {
		head = tail = 0 ;
		clear ( d , -1 ) ;
		clear ( num , 0 ) ;
		Q[tail ++] = t ;
		d[t] = 0 ;
		num[d[t]] = 1 ;
		while ( head != tail ) {
			int u = Q[head ++] ;
			for ( int i = H[u] ; ~i ; i = E[i].n ) {
				int v = E[i].v ;
				if ( ~d[v] )
					continue ;
				d[v] = d[u] + 1 ;
				num[d[v]] ++ ;
				Q[tail ++] = v ;
			}
		}
	}
	
	int ISAP () {
		copy ( cur , H ) ;
		rev_bfs () ;
		int u = pre[s] = s , i ;
		while ( d[s] < nv ) {
			if ( u == t ) {
				int f = INF , pos ;
				for ( i = s ; i != t ; i = E[cur[i]].v )
					if ( f > E[cur[i]].c ) {
						f = E[cur[i]].c ;
						pos = i ;
					}
				for ( i = s ; i != t ; i = E[cur[i]].v ) {
					E[cur[i]].c -= f ;
					E[cur[i] ^ 1].c += f ;
				}
				u = pos ;
				flow += f ;
			}
			for ( i = cur[u] ; ~i ; i = E[i].n )
				if ( E[i].c && d[u] == d[E[i].v] + 1 )
					break ;
			if ( ~i ) {
				cur[u] = i ;
				pre[E[i].v] = u ;
				u = E[i].v ;
			}
			else {
				if ( 0 == ( -- num[d[u]] ) )
					break ;
				int mmin = nv ;
				for ( i = H[u] ; ~i ; i = E[i].n )
					if ( E[i].c && mmin > d[E[i].v] ) {
						mmin = d[E[i].v] ;
						cur[u] = i ;
					}
				d[u] = mmin + 1 ;
				num[d[u]] ++ ;
				u = pre[u] ;
			}
		}
		return flow ;
	}
	
	int unique ( int a[] , int n ) {
		int cnt = 1 ;
		sort ( a + 1 , a + n + 1 ) ;
		REPF ( i , 2 , n )
			if ( a[i] != a[cnt] )
				a[++ cnt] = a[i] ;
		return cnt ;
	}
	
	int binary_search ( int x , int l , int r ) {
		while ( l < r ) {
			int m = ( l + r ) >> 1 ;
			if ( a[m] >= x )
				r = m ;
			else
				l = m + 1 ;
		}
		return l ;
	}
	
	void input () {
		m = 0 ;
		scanf ( "%d" , &n ) ;
		REPF ( i , 1 , n ) {
			A[i].input () ;
			a[++ m] = A[i].l ;
			a[++ m] = ++ A[i].r ;
			//[ , ]转化为[ , )
		}
		m = unique ( a , m ) ;
	}
	
	void build () {
		REPF ( i , 2 , m ) {
			addedge ( n + i , t , a[i] - a[i - 1] ) ;
			REPF ( j , 1 , n )
				if ( A[j].l <= a[i - 1] && a[i] <= A[j].r )
					addedge ( j , n + i , 1 ) ;
		}
	}
	
	void solve () {
		init () ;
		input () ;
		s = 0 ;
		t = m + n + 1 ;
		nv = t + 1 ;
		build () ;
		int cnt = cntE ;
		flow = 0 ;
		REPV ( i , n , 1 ) {
			addedge ( s , i , 1 ) ;
			ISAP () ;
		}
		printf ( "%d\n" , flow ) ;
		int flag = 0 ;
		for ( int i = H[0] ; ~i ; i = E[i].n ) {
			if ( !E[i].c ) {
				if ( flag )
					printf ( " " ) ;
				flag = 1 ;
				printf ( "%d" , E[i].v ) ;
			}
		}
		printf ( "\n" ) ;
		
	}
} ;

NetWork nw ;

int main () {
	int T ;
	scanf ( "%d" , &T ) ;
	while ( T -- )
		nw.solve () ;
	return 0 ;
}