Poj Solution 1944

```http://poj.org/problem?id=1944

#include <stdio.h>
#include <algorithm>
#include <memory.h>
using namespace std;
int t[4100];
int sum[4100];
void insert( int l, int r, int a, int b, int s, bool del )
{
int c = (l+r)/2;
if( l>=r || l>=b || r<=a ) return;
if( a<=l && r<=b )
{
if(del)
{
t[s]--;
if( !t[s] ) sum[s] = (r-l==1) ? 0 : ( sum[s*2+1] + sum[s*2+2] );
}
else
{
t[s]++;
sum[s] = r - l;
}
return;
}
insert( l, c, a, b, s*2+1, del );
insert( c, r, a, b, s*2+2, del );
if( !t[s] )    sum[s] = sum[s*2+1] + sum[s*2+2];
}
typedef pair<int,int> seg;
seg h[20010];
int n, p;
void init()
{
int i;
scanf( "%d %d", &n, &p );
memset( t, 0, 4100*sizeof(int) );
memset( sum, 0, 4100*sizeof(int) );
for( i=0; i<p; i++ )
{
scanf( "%d %d", &h[i].first, &h[i].second );

h[i].first--, h[i].second--;

if( h[i].first > h[i].second ) swap( h[i].first, h[i].second );

insert( 0, 2048, h[i].first, h[i].second, 0, false );

h[i+p].first = h[i].second;
h[i+p].second = h[i].first+n;
}

p*=2;

sort( h, h+p );

}
int main()
{
int i, ans, j;
init();
ans = 10000;
for( i=0, j=0 ; i<n; i++ )
{
if( sum[0] < ans ) ans = sum[0];
while( j < p && h[j].first <= i )
{
insert( 0, 2048, h[j].first, h[j].second, 0, true );
insert( 0, 2048, h[j].second, h[j].first+n, 0, false );
j++;
}
}
printf( "%dn", ans );
return 0;
}

```
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