poj 1887 Testing the CATCHER(最長遞增子序列)

題意：給你一些數找到最長遞減子序列

思路：跟nyoj 214  差不多，只不過這兒是單調遞減

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;

#define maxN 40000
int f[maxN],so[maxN],key[maxN],n;

bool cmp(int &x,int &y)
{
	return x>y;
}
int low_bun(int l,int r,int x)
{
	if(l==r)return l;
	int mid=(l+r)/2;
	if(so[mid]<=x)return low_bun(l,mid,x);
	else return low_bun(mid+1,r,x);
}
int low(int x){return x&(-x);}
int get_max(int x)
{
	int maxx=0;
	while(x!=0){
		maxx=max(maxx,key[x]);
		x-=low(x);
	}
	return maxx;
}
void update(int x)
{
	int m=x;
	while(x<n+1){
		if(key[m]>=key[x])
			key[x]=key[m];
		else
			break;
		x+=low(x);
	}
}

int main()
{
	int k=1;
	while(scanf("%d",&f[1])&&f[1]!=-1)
	{
		memset(key,0,sizeof(key));
		n=2;
		so[0]=0x0fffff,so[1]=f[1];
		while(scanf("%d",&f[n])&&f[n]!=-1)
			so[n]=f[n++];
		sort(so,so+n,cmp);
		int maxx=0;
		for(int i=1;i<n;i++)
		{
			int m=low_bun(1,n-1,f[i]);
			key[m]=get_max(m-1)+1;
			maxx=max(maxx,key[m]);
			update(m);
		}
		printf("Test #%d:\n  maximum possible interceptions: %d\n\n",k++,maxx);
	}
}