aoj1313（*simpson公式求三维体积）

/*
translation:
三维坐标系上有2个棱柱，分别平行与Z轴以及Y轴，现在分别给出它们在XY平面和XZ平面上的底面顶点，求两个棱柱相交部分的体积。
solution:
主要思路是不难，但前提得知道simpson公式。利用该公式即可将积分公式化为普通的公式。然后微积分求出来的既是体积。
*/
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>

using namespace std;
const int maxn = 200 + 5;
const double INF = 0x3f3f3f3f * 1.0;

int X1[maxn], Y1[maxn], m, n;
int X2[maxn], Z2[maxn];
vector<int> xv;

double getWidth(int* X, int* Y, int n, double x)
{
double lb = INF, ub = -INF;
for(int i = 0; i < n; i++) {
double ax = X[i], ay = Y[i];
double bx = X[(i+1)%n], by = Y[(i+1)%n];
if((ax - x) * (bx - x) <= 0 && ax != bx) {
double y = (by - ay) / (bx - ax) * (x - ax) + ay;
lb = min(lb, y);	ub = max(ub, y);
}
}
return max(0.0, ub-lb);
}

int main()
{
//freopen("in.txt", "r", stdin);
while(~scanf("%d%d", &m, &n)) {
if(m + n == 0)	break;
for(int i = 0; i < m; i++) scanf("%d%d", &X1[i], &Y1[i]);
for(int i = 0; i < n; i++) scanf("%d%d", &X2[i], &Z2[i]);

xv.clear();
for(int i = 0; i < m; i++)	xv.push_back(X1[i]);
for(int i = 0; i < n; i++)	xv.push_back(X2[i]);

int min1 = *min_element(X1, X1 + m), max1 = *max_element(X1, X1 + m);
int min2 = *min_element(X2, X2 + n), max2 = *max_element(X2, X2 + n);

sort(xv.begin(), xv.end());

double ans = 0;
for(int i = 0; i < xv.size()-1; i++) {
double a = xv[i], b = xv[i+1], c = (a + b) / 2;
if(c >= min1 && c <= max1 && c >= min2 && c <= max2) {
double fa = getWidth(X1, Y1, m, a) * getWidth(X2, Z2, n, a);
double fb = getWidth(X1, Y1, m, b) * getWidth(X2, Z2, n, b);
double fc = getWidth(X1, Y1, m, c) * getWidth(X2, Z2, n, c);
ans += (b - a) / 6 * (fa + 4 * fc + fb);
}
}
printf("%.10f\n", ans);
}
return 0;
}