zoj 3913 Bob wants to pour water(二分)

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题目链接：zoj 3913 Bob wants to pour water

解题思路

二分高度，判断容量。

代码

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int maxn = 1e5 + 5;const double eps = 1e-8;const double pi = 4 * atan(1);inline int dcmp(double x) {    if (fabs(x) < eps) return 0;    return x < 0 ? -1 : 1;}struct Rectangle {    double z, w, l, h;    void read () { scanf("%lf%lf%lf%lf", &z, &w, &l, &h); }    double volume () { return w * l * h; }    double contain(double H) {        double dn = z-h/2, up = min(z+h/2, H);        return w * l * max(0.0, up-dn);    }}R[maxn];struct Circle {    double z, r;    void read () { scanf("%lf%lf", &z, &r); }    double volume () { return pow(r, 3) * pi * 4 / 3; }    double lost(double H) { return pi * H * H * (r - H / 3); }    double contain(double H) {        if (dcmp(H-(z+r)) >= 0) return volume();        if (dcmp(H-(z-r)) <= 0) return 0;        if (dcmp(H-z) >= 0) return volume()-lost(z+r-H);        else return lost(H-(z-r));    }}C[maxn];int M, N;double W, L, V;double getVolume(double H) {    double ret = H * W * L;    for (int i = 0; i < M; i++)        ret -= R[i].contain(H);    for (int i = 0; i < N; i++)        ret -= C[i].contain(H);    return ret;}double solve () {    double l = 0, r = V / (W*L);    for (int i = 0; i < M; i++) r += R[i].volume()/(W*L);    for (int i = 0; i < N; i++) r += C[i].volume()/(W*L);    //while (r-l > 1e-4) {    for (int i = 0; i < 200; i++) {        double mid = (l + r) / 2;        double vol = getVolume(mid);        if (dcmp(vol-V) > 0) r = mid;        else l = mid;    }    return l;}int main () {    int cas;    scanf("%d", &cas);    while (cas--) {        scanf("%lf%lf%lf%d%d", &W, &L, &V, &M, &N);        for (int i = 0; i < M; i++) R[i].read();        for (int i = 0; i < N; i++) C[i].read();        printf("%.6lf\n", solve());    }    return 0;}

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