Codeforces 138C(区间更新+离散化)

题意：有n棵树在水平线上，给出每棵树的坐标和高度，然后向左倒的概率和向右倒的概率，和为1，然后给出了m个蘑菇的位置，每个蘑菇都有一个魔法值，如果蘑菇被压死了，也就是在某棵树[a[i] - h[i], a[i]) 或 (a[i], a[i] + h[i]]范围内，魔法值就没有了，只有生存下来的蘑菇才有魔法值，问生存下来的蘑菇的魔法值的期望。

题解：可以看到n和m的范围是1e5，而坐标范围是1e9，所以肯定要离散化，然后更新每个区间的概率值，单点查询每个蘑菇所在区间的概率值乘其魔法值。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
const int N = 100005;
int n, m, a
, h
, b
, z
, c[N << 2];
double tree[N << 4], flag[N << 4], pl
, pr
;
map<int, int> mp;

void pushdown(int k) {
if (flag[k]) {
tree[k * 2] *= tree[k];
tree[k * 2 + 1] *= tree[k];
flag[k * 2] = flag[k * 2 + 1] = 1;
tree[k] = 1.0;
flag[k] = 0;
}
}

void build(int k, int left, int right) {
flag[k] = 0;
tree[k] = 1.0;
if (left != right) {
int mid = (left + right) / 2;
build(k * 2, left, mid);
build(k * 2 + 1, mid + 1, right);
}
}

void modify(int k, int left, int right, int l1, int r1, double x) {
if (l1 <= left && right <= r1) {
tree[k] *= x;
flag[k] = 1;
return;
}
pushdown(k);
int mid = (left + right) / 2;
if (l1 <= mid)
modify(k * 2, left, mid, l1, r1, x);
if (r1 > mid)
modify(k * 2 + 1, mid + 1, right, l1, r1, x);
}

double query(int k, int left, int right, int pos) {
if (left == right)
return tree[k];
pushdown(k);
int mid = (left + right) / 2;
if (pos <= mid)
return query(k * 2, left, mid, pos);
else
return query(k * 2 + 1, mid + 1, right, pos);
}

int main() {
scanf("%d%d", &n, &m);
mp.clear();
int cnt = 0;
for (int i = 1; i <= n; i++) {
scanf("%d%d%lf%lf", &a[i], &h[i], &pl[i], &pr[i]);
pl[i] /= 100.0, pr[i] /= 100.0;
c[++cnt] = a[i];
c[++cnt] = a[i] - h[i];
c[++cnt] = a[i] + h[i];
}
for (int i = 1; i <= m; i++) {
scanf("%d%d", &b[i], &z[i]);
c[++cnt] = b[i];
}
sort(c + 1, c + 1 + cnt);
cnt = unique(c + 1, c + 1 + cnt) - (c + 1);
for (int i = 1; i <= cnt; i++)
mp[c[i]] = i;
build(1, 1, cnt);
for (int i = 1; i <= n; i++) {
modify(1, 1, cnt, mp[a[i] - h[i]], mp[a[i]] - 1, 1.0 - pl[i]);
modify(1, 1, cnt, mp[a[i]] + 1, mp[a[i] + h[i]], 1.0 - pr[i]);
}
double res = 0;
for (int i = 1; i <= m; i++)
res += z[i] * query(1, 1, cnt, mp[b[i]]);
printf("%lf\n", res);
return 0;
}