HDU 5481：Desiderium（线段树）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5481

题意：给定一个n(1≤n≤100000)(1 \le n \le 100000)条线段的集合，要求求出该集合所有子集的线段并集的长度的和。

分析：首先将线段进行离散化处理。利用线段树处理出每段线段被覆盖的次数m，则在最终的长度和里面，该段线段被计算的次数为2n−2n−m2^n-2^{n-m}。最后累加即可。

代码：

#include <iostream>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <string>
#include <cstring>

using namespace std;

#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1

const int maxn = 100000 + 5, INF = 2000005;
const long long mod = 1e9 + 7;

int T, n, num, x;
int l[maxn], r[maxn], a[maxn << 1];
int col[maxn << 3];
int sum[maxn << 3];
long long tmp, ans;

void PushUp(int rt)
{
sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}

void PushDown(int rt, int m)
{
if (col[rt])
{
col[rt << 1] += col[rt];
col[rt << 1 | 1] += col[rt];
sum[rt << 1] += ((m - (m >> 1)) * col[rt]);
sum[rt << 1 | 1] += ((m >> 1) * col[rt]);
col[rt] = 0;
}
}

void build(int l, int r, int rt)
{
col[rt] = 0;
sum[rt] = 0;
if (l == r)
return;
int m = (l + r) >> 1;
build(lson);
build(rson);
PushUp(rt);
}

void update(int L, int R, int c, int l, int r, int rt)
{
if (L <= l && r <= R)
{
col[rt] += c;
sum[rt] += c * (r - l + 1);
return;
}
PushDown(rt , r - l + 1);
int m = (l + r) >> 1;
if (L <= m)
update(L, R, c, lson);
if (R > m)
update(L, R, c, rson);
PushUp(rt);
}

int query(int L, int R, int l, int r, int rt)
{
if (L <= l && r <= R)
return sum[rt];
PushDown(rt , r - l + 1);
int m = (l + r) >> 1;
int ret = 0;
if (L <= m)
ret += query(L , R , lson);
if (R > m)
ret += query(L , R , rson);
return ret;
}

int ID(int x)
{
return lower_bound(a, a + num, x) - a;
}

long long quick_power(int x, int y)
{
long long res = 1, base = x;
while (y)
{
if (y & 1)
res = (res * base) % mod;
base = (base * base) % mod;
y >>= 1;
}
return res % mod;
}

int main()
{
scanf("%d", &T);
for (int C = 0; C < T; ++C)
{
scanf("%d", &n);
for (int i = 0; i < n; ++i)
{
scanf("%d%d", &l[i], &r[i]);
a[i << 1] = l[i];
a[i << 1 | 1] = r[i];
}
sort(a, a + n + n);
num = unique(a, a + n + n) - a;
build(1, num, 1);
for (int i = 0; i < n; ++i)
update(ID(l[i]) + 1, ID(r[i]), 1, 1, num, 1);
tmp = quick_power(2, n);
ans = 0;
for (int i = 0; i < num - 1; ++i)
{
x = query(i + 1, i + 1, 1, num , 1);
ans = (ans + ((((((tmp - quick_power(2, n - x)) % mod) + mod) % mod)) * ((a[i + 1] - a[i]) % mod)) % mod) % mod;
}
printf("%I64d\n", ans);
}
return 0;
}