zoj 1025

這題剛開始直接dp求最長非遞減子序列，再加上next數組，於是WA。結合網上的代碼，才發現，其實就是求最少組的非遞減子序列，直接從小到大掃描就可以了。自己還是太水了，後悔大學不努力呀！

/*
 *	zoj_1025,就是求最少組的非遞減子序列，用貪心就可以，不用dp求最長非遞減子序列
 */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define N	5001
struct Stick {
	int len, w;
};
Stick stick[N];
short flag[N];

int cmp( const Stick& a, const Stick& b ) {
	return ( a.len < b.len || ( a.len == b.len && a.w < b.w ) );
}
int main()
{
	int i, j, tmp, t, n, ans;
	scanf("%d", &t );
	while ( t -- ) {
		scanf("%d", &n);
		for ( i = 0; i < n; ++ i )
			scanf("%d%d", &stick[i].len, &stick[i].w );
		sort( stick, stick+n, cmp );
		memset( flag, 0, sizeof(flag) );
		ans = 0;
		for ( i = 0; i < n; ++ i ) {
			if ( 0 == flag[i] ) {
				++ ans;
				tmp = stick[i].w;
				for ( j = i+1; j < n; ++ j ) {
					if ( 0 == flag[j] && stick[j].w >= tmp ) {
						flag[j] = 1;
						tmp = stick[j].w;
					}
				}
			}
		}
		printf("%d\n", ans );
	}
	return 0;
}