題意：有n棵樹在水平線上，給出每棵樹的坐標和高度，然后向左倒的概率和向右倒的概率，和為1，然后給出了m個蘑菇的位置，每一個蘑菇都有一個魔法值，假設蘑菇被壓死了，也就是在某棵樹[a[i] - h[i], a[i]) 或 (a[i], a[i] + h[i]]范圍內。魔法值就沒有了。僅僅有生存下來的蘑菇才有魔法值，問生存下來的蘑菇的魔法值的期望。
題解：能夠看到n和m的范圍是1e5。而坐標范圍是1e9。所以肯定要離散化，然后更新每一個區間的概率值，單點查詢每一個蘑菇所在區間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[N], h[N], b[N], z[N], c[N << 2]; double tree[N << 4], flag[N << 4], pl[N], pr[N]; 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);elsereturn 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; }

轉載于:https://www.cnblogs.com/llguanli/p/7401510.html

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