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  • Codeforces 138C Mushroom Gnomes

    Mushroom Gnomes - 2

    感觉没啥东西, 用线段树算算每个被覆盖的概率, 坑点是有很多个在同一个点。

    #include<bits/stdc++.h>
    #define LL long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PLL pair<LL, LL>
    #define PLI pair<LL, int>
    #define PII pair<int, int>
    #define SZ(x) ((int)x.size())
    #define ull unsigned long long
    
    using namespace std;
    
    const int N = 4e5 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 1e9 + 7;
    const double eps = 1e-8;
    const double PI = acos(-1);
    
    int A[N], h[N], b[N], X[N], n, m, tot;
    double pl[N], pr[N], z[N], val[N];
    
    #define lson l, mid, rt << 1
    #define rson mid + 1, r, rt << 1 | 1
    double a[N << 2], lazy[N << 2];
    void push(int rt) {
        if(fabs(lazy[rt] - 1) > eps) {
            a[rt << 1] *= lazy[rt];
            a[rt << 1 | 1] *= lazy[rt];
            lazy[rt << 1] *= lazy[rt];
            lazy[rt << 1 | 1] *= lazy[rt];
            lazy[rt] = 1;
        }
    }
    void build(int l, int r, int rt) {
        lazy[rt] = 1;
        if(l == r) {
            a[rt] = 1;
            return;
        }
        int mid = l + r >> 1;
        build(lson); build(rson);
    }
    void update(int L, int R, double val, int l, int r, int rt) {
        if(L > R) return;
        if(l >= L && r <= R) {
            a[rt] *= val;
            lazy[rt] *= val;
            return;
        }
        int mid = l + r >> 1;
        push(rt);
        if(L <= mid) update(L, R, val, lson);
        if(R > mid)  update(L, R, val, rson);
    }
    double query(int p, int l, int r, int rt) {
        if(l == r) return a[rt];
        int mid = l + r >> 1;
        push(rt);
        if(p <= mid) return query(p, lson);
        else return query(p, rson);
    }
    
    int getId(int x) {
        return lower_bound(X, X + tot, x) - X + 1;
    }
    
    int main() {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++) {
            scanf("%d%d", &A[i], &h[i]);
            scanf("%lf%lf", &pl[i], &pr[i]);
            pl[i] /= 100; pr[i] /= 100;
            X[tot++] = A[i];
            X[tot++] = A[i] + h[i];
            X[tot++] = A[i] - h[i];
        }
        for(int i = 1; i <= m; i++) {
            scanf("%d%lf", &b[i], &z[i]);
            X[tot++] = b[i];
        }
        sort(X, X + tot);
        tot = unique(X, X + tot) - X;
        build(1, tot, 1);
        for(int i = 1; i <= m; i++) {
            val[getId(b[i])] += z[i];
        }
        for(int i = 1; i <= tot; i++) {
            update(i, i, val[i], 1, tot, 1);
        }
        for(int i = 1; i <= n; i++) {
            update(getId(A[i]) + 1, getId(A[i] + h[i]), 1 - pr[i], 1, tot, 1);
            update(getId(A[i] - h[i]), getId(A[i]) - 1, 1 - pl[i], 1, tot, 1);
        }
        double ans = 0;
        for(int i = 1; i <= tot; i++) {
            ans += query(i, 1, tot, 1);
        }
        printf("%.12f
    ", ans);
        return 0;
    }
    
    /*
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/10487094.html
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