HDU 1556 Color the ball

Description

N个气球排成一排，从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球a开始到气球b依次给每个气球涂一次颜色。但是N次以后lele已经忘记了第I个气球已经涂过几次颜色了，你能帮他算出每个气球被涂过几次颜色吗？

 

Input

每个测试实例第一行为一个整数N,(N <= 100000).接下来的N行，每行包括2个整数a b(1 <= a <= b <= N)。 
当N = 0，输入结束。

 

Output

每个测试实例输出一行，包括N个整数，第I个数代表第I个气球总共被涂色的次数。

 

Sample Input

     
     
     
     

      
      
      
      3
1 1
2 2
3 3
3
1 1
1 2
1 3
0 
     
     
     
     

 

Sample Output

     
     
     
     

      
      
      
      1 1 1
3 2 1 
     
     
     
     

 

简单的区间更新单点询问，拿来用splay练练延迟更新

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<functional>
using namespace std;
typedef unsigned long long ull;
typedef long long LL;
int n, m, l, r, c, root;
char s[10];

struct Splays
{
	const static int maxn = 1e5 + 10;
	const static int INF = 0x7FFFFFFF;
	int ch[maxn][2], F[maxn], U[maxn], C[maxn], A[maxn], sz, G[maxn];
	LL S[maxn];
	int Node(int f, int u, int c) { C[sz] = 1; S[sz] = A[sz] = c; G[sz] = ch[sz][0] = ch[sz][1] = 0; F[sz] = f; U[sz] = u; return sz++; }
	void clear(){ sz = 1; ch[0][0] = ch[0][1] = C[0] = A[0] = U[0] = F[0] = S[0] = G[0] = 0; }
	void Pushdown(int x)
	{
		G[ch[x][0]] += G[x];
		G[ch[x][1]] += G[x];
		S[ch[x][0]] += (LL)G[x] * C[ch[x][0]];
		S[ch[x][1]] += (LL)G[x] * C[ch[x][1]];
		A[x] += G[x];	G[x] = 0;
	}
	void rotate(int x, int k)
	{
		int y = F[x];	Pushdown(y);	Pushdown(x);
		ch[y][!k] = ch[x][k]; F[ch[x][k]] = y; 
		if (F[y]) ch[F[y]][y == ch[F[y]][1]] = x;
		F[x] = F[y];    F[y] = x;	ch[x][k] = y;		
		C[x] = C[y];	C[y] = C[ch[y][0]] + C[ch[y][1]] + 1;
		S[x] = S[y];	S[y] = S[ch[y][0]] + S[ch[y][1]] + A[y];
	}
	void Splay(int x, int r)
	{
		for (int fa = F[r]; F[x] != fa;)
		{
			if (F[F[x]] == fa) { rotate(x, x == ch[F[x]][0]); return; }
			int y = x == ch[F[x]][0], z = F[x] == ch[F[F[x]]][0];
			y^z ? (rotate(x, y), rotate(x, z)) : (rotate(F[x], z), rotate(x, y));
		}
	}
	bool insert(int &x, int u, int v)
	{
		if (!x) { x = Node(0, u, v); return false; }
		int now = 0;
		for (int i = x; i&&!now; C[i]++, S[i] += v, i = ch[i][U[i] < u])
		{
			//Pushdown(i);
			if (u == U[i]) { A[i] += v; Splay(i, x); x = i; return true; }
			if (!ch[i][U[i] < u]) now = ch[i][U[i] < u] = Node(i, u, v);
		}
		Splay(now, x);  x = now; return false;
	}
	void add(int &x, int l, int r, int c)
	{
		insert(x, l - 1, 0);	
		insert(ch[x][1], r + 1, 0);
		G[ch[ch[x][1]][0]] += c;
		S[ch[ch[x][1]][0]] += (LL)c*C[ch[ch[x][1]][0]];
		S[ch[x][1]] += (LL)c*C[ch[ch[x][1]][0]];
		S[x] += (LL)c*C[ch[ch[x][1]][0]];
	}
	void find(int &x, int l, int r)
	{
		insert(x, l - 1, 0);
		insert(ch[x][1], r + 1, 0);
		printf("%d\n", S[ch[ch[x][1]][0]]);
	}
	void find(int &x, int l)
	{
		insert(x, l, 0);
		printf("%d", A[x]);
	}
}solve;

int main()
{
	while (scanf("%d", &n) != EOF, n)
	{
		solve.clear();	root = 0;
		for (int i = 1; i <= n; i++) solve.insert(root, i, 0);
		solve.insert(root, 0, 0);	solve.insert(root, n + 1, 0);
		for (int i = 1; i <= n; i++)
		{
			scanf("%d%d", &l, &r);
			solve.add(root, l, r, 1);
		}
		for (int i = 1; i <= n; i++) solve.find(root, i), printf("%s", i == n ? "\n" : " ");
	}
	return 0;
}