Sichuan University Programming Contest 2013 Final （singal）E题

题目链接：http://cstest.scu.edu.cn/soj/problem.action?id=4285

这道题涉及到动态加入和删除以及求第k大数，典型用树状数组和线段树，但是比赛时真心没有读懂题目。由于题目中weight较大，肯定要离散化，但是数是动态加入的。所以肯定先离线处理，再离散化。贴下代码：

#include <stdio.h>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <string.h>
using namespace std;
const int MAX = (5000+10) * 1000;
int n;
int  brick[1001][5001], m[1001];
map<int ,int> mapii;
set<int> a;
int mapxx[MAX];
int c[MAX], weight, num;
bool flag[MAX];
int lowbit(int x)
{
	return x&(-x);
}

void update(int x, int v)
{
	while (x<MAX)
	{
		c[x] += v;
		x += lowbit(x);
	}
}

int getsum(int x)
{
	int sum = 0;
	while (x)
	{
		sum += c[x];
		x -= lowbit(x);
	}
	return sum;
}

int find(int x)
{
	int l = 1, r = MAX, mid;
	while (l<r)
	{
		mid = (l + r) >> 1;
		if (getsum(mid)<x)
		{
			l = mid+1;
		}
		else
		{
			r = mid;
		}
	}
	return l;
}
int main()
{
	int t, i, j;
	scanf("%d",&t);
	while (t--)
	{
		num=0;
		a.clear();
		memset(c, 0, sizeof(c));
		memset(flag, 0, sizeof(flag));
		scanf("%d", &n);
		for (i=0; i<n; i++)
		{
			scanf("%d", &m[i]);
			for (j=0; j<m[i]; j++)
			{
				scanf("%d", &weight);
				brick[i][j] = weight;
				a.insert(weight);
			}
		}
		set<int>::iterator id;
		for (i=1, id = a.begin(); id!=a.end(); i++, id++)
		{
			mapii[*id] = i;
			mapxx[i] = *id;
		}
		for (i=0; i<n; i++)
		{
			for (j=0; j<m[i]; j++)
			{
				if (!flag[mapii[brick[i][j]]])
				{
					num++;
					update(mapii[brick[i][j]], 1);
					flag[mapii[brick[i][j]]] = true;
				}
			}
			if (num < 2)
			{
				puts("-1 -1");
				continue;
			}
			int t1 = find(2);
			int t2 = find(num-1);
			printf("%d %d\n", mapxx[t1], mapxx[t2]);
			update(t1, -1);
			update(t2, -1);
			flag[t1] = false;
			flag[t2] = false;
			num-=2;
		}	
	}
}