aoj1313（*simpson公式求三维体积）

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/*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;}
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