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;
}