题目连接：
codeforces 4D
题目大意：
给出n个信封，，这n个信封有长和宽，给出卡片的尺寸，求取能够装入卡片的最长的序列，序列满足后一个的长和宽一定大于前一个，求最长的这个序列的长度，并且给出一组可行解。
题目分析：答案递归地倒序输出即可。 AC代码：;int n,w,h;int dp[MAX];int back[MAX];struct Node{int w,h,x;bool operator < ( const Node & a ) const{if ( w == a.w ) return h < a.h;return w < a.w;}}a[MAX];void print ( int x ){if ( a[x]. w <= w || a[x].h <= h ) return;print ( back[x] );printf ( “%d ” , a[x].x );}int main ( ){while ( ~scanf ( “%d%d%d” , &n , &w , &h )){for ( int i = 1 ; i <= n ; i++ ){scanf ( “%d%d” , &a[i].w , &a[i].h );a[i].x = i;}dp[0] = -1;sort ( a+1 , a+n+1 );for ( int i = 1 ; i <= n ; i++ ){int temp = -1,id = 0;for ( int j = 0 ; j < i ; j++ )if ( a[j].w < a[i].w && a[j].h < a[i].h && a[j].w > w && a[j].h > h )if ( temp < dp[j] ){temp = dp[j];id = j;}if ( a[i].w > w && a[i].h > h ) dp[i] = 1 , back[i] = 0;else dp[i] = -1;if ( temp+1 > dp[i] ){dp[i] = temp+1;back[i] = id;}}int maxn = 0;for ( int i = 1 ; i <= n ; i++ )maxn = max ( maxn , dp[i] );if ( maxn == 0 ){puts ( “0” );continue;}printf ( “%d\n” , maxn );for ( int i = 1 ; i <= n ; i++ )if ( maxn == dp[i] ){print ( i );puts (“”);break;}}}
，再回头，便生出无限羁绊。那是彼此的刺在对方心里留下的痕迹，
