题意:有一段区间,n条限制区间以及一个常数k。现在要求长度>=k的限制区间至少有k个点被选中,长度小于k的区间所有点都要被选中,问最少需要选中多少个点才能满足限制要求,并升序输出其中一个可行解。
思路:大量不等式关系求最小值,直接想到差分约束。建图后spfa()一个最长路,然后遍历区间判断dis[i]>dis[i-1],是则表明被选中反之不选中。对于区间可能涉及负数,可以进行移位操作所有点向右移动1-Min个单位。(Min,Max分别指所有区间边界的最小值和最大值)
C++代码:
#include<map>
#include<set>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 20010;
const int maxm = 50010;
const int inf = 0x3f3f3f3f;
int k,n,tol,head[maxn];
struct edge
{
int to,cost,next;
}es[maxm];
void init()
{
tol = 0;
memset ( head , -1 , sizeof(head) );
}
void addedge( int u , int v , int w )
{
es[tol].to = v;
es[tol].cost = w;
es[tol].next = head[u];
head[u] = tol++;
}
int a[maxn];
int b[maxn];
int vis[maxn];
int dis[maxn];
void spfa()
{
memset ( vis , 0 , sizeof(vis) );
memset ( dis , -inf , sizeof(dis) );
queue<int>Q;
Q.push(0);
dis[0] = 0;
vis[0] = 1;
while ( !Q.empty() )
{
int u = Q.front();
Q.pop();
vis[u] = 0;
for ( int i=head[u] ; i!=-1 ; i=es[i].next )
{
int v = es[i].to,w = es[i].cost;
if ( dis[v]<dis[u]+w )
{
dis[v] = dis[u]+w;
if ( !vis[v] )
{
vis[v] = 1;
Q.push(v);
}
}
}
}
}
int main()
{
while ( scanf ( "%d%d" , &k , &n )==2 )
{
init();
int Min = inf,Max = -inf;
for ( int i=1 ; i<=n ; i++ )
{
scanf ( "%d%d" , &a[i] , &b[i] );
if ( a[i]>b[i] ) swap( a[i] , b[i] );
Min = min( Min , a[i] );
Max = max( Max , b[i] );
}
for ( int i=1 ; i<=n ; i++ )
{
a[i] += ( 1-Min );
b[i] += ( 1-Min );
if ( b[i]-a[i]+1>=k )
addedge( a[i]-1 , b[i] , k );
else
{
int c = b[i]-a[i]+1;
addedge( a[i]-1 , b[i] , c );
addedge( b[i] , a[i]-1 , -c );
}
}
for ( int i=1 ; i<=Max+(1-Min) ; i++ )
addedge( i-1 , i , 0 ),addedge( i , i-1 , -1 );
spfa();
printf ( "%d\n" , dis[Max+(1-Min)] );
for ( int i=1 ; i<=Max+(1-Min) ; i++ )
if ( dis[i]>dis[i-1] )
printf ( "%d\n" , i+Min-1 );
}
return 0;
}