Zoj 1025-wooden sticks

Question: calculate the maximum number of subsequences that do not increase in a sequence.

Analysis: DP, Lis. The minimum number of non-ascending subsequences in a sequence is its maximum ascending subsequence length.

Proof: First, find the Maximum ascending sub-sequence of the string, then each element must belong to a different non-descending string;

If the first largest ascending subsequence is used, each element must be the maximum value in the set;

The maximum values can be used to determine a substring instead of a substring, which corresponds to a set. Therefore, the lower bound is | Lis |;

Assume that a set P exists. The maximum value p is not in LIS and does not belong to the preceding set;

He determines a sequence without going up, and sets the maximum value of the set before him to Lis (I );

Then there is Lis (I)> = P (otherwise P is Lis (I + 1); replace Lis (I) with P in the collection,

For a part smaller than P, we obtain the set determined by the new Lis (I) and a new independent set P1;

If you set the maximum P1 value to P1, there are two possibilities:

1. Lis (I-1) <P1 <Lis (I) can constitute a longer Lis, then the result is still | Lis |, prove;

2. Lis (I-1)> = p1 repeat the above operation and keep getting the PJ set until 1 is set up or I = 1;

If I is 1, pj <Lis (1) goes to the top of the longer Lis; otherwise, 1 is true.

Note: The monotonous queue can be an O (N ^ 2) algorithm and optimized to O (nlogn ). )

#include <stdio.h>#include <stdlib.h>typedef struct node{    int l,w;}stick;int BS( int h, int key, int *MUQ ){    int l = 0,m;    while ( l < h ) {        m = (l+h)/2;         if ( key >= MUQ[ m ] )            h = m;        else l = m+1;    }    return h;}int cmp( const void* a, const void* b ){    stick *p = (stick *)a;    stick *q = (stick *)b;    if ( p->l == q->l )        return p->w - q->w;    return p->l - q->l;}int main(){    stick Stick[ 5001 ];    int   MUQ[ 5001 ];        int t,n,i,tail,j;    while ( scanf("%d",&t) != EOF )    while ( t -- ) {        scanf("%d",&n);        for ( i = 1 ; i <= n ; ++ i  )            scanf("%d%d",&Stick[ i ].l,&Stick[ i ].w);                qsort( &Stick[ 1 ], n, sizeof( stick ), cmp );        tail = 0;        MUQ[ 0 ] = Stick[ 1 ].w;                for ( i = 2 ; i <= n ; ++ i )             if ( Stick[ i ].w < MUQ[ tail ] )                 MUQ[ ++ tail ] = Stick[ i ].w;            else MUQ[ BS( tail, Stick[ i ].w, MUQ ) ] = Stick[ i ].w;                printf("%d\n",tail+1);    }    return 0;}

Zoj 1025-wooden sticks