挑战程序竞赛系列（95）：3.6数值积分（1）

挑战程序竞赛系列（95）：3.6数值积分（1）

传送门：AOJ 1313: Intersection of Two Prisms

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题意：

有一个侧棱与Z轴平行的棱柱P1和一个侧棱与y轴平行的棱柱P2。它们都向两端无限延伸，底面分别是包含M个顶点和N个顶点的凸多边形，其中第i个顶点的坐标分别是（X1, Y1）和（X2, Y2）。请计算这两个棱柱公共部分的体积。

按照x轴进行切片，求出每一个瞬间的面积（积分），所以只需要知道给定x，求出f(x)即可。书中给出了积分的近似公式：

∫baf(x)dx=b−a6(f(a)+4f(a+b2)+f(b))

代码如下：

import java.io.BufferedReader;import java.io.File;import java.io.FileInputStream;import java.io.IOException;import java.io.InputStream;import java.io.InputStreamReader;import java.io.PrintWriter;import java.util.ArrayList;import java.util.Collections;import java.util.List;import java.util.StringTokenizer;public class Main{    String INPUT = "./data/judge/201710/A1313.txt";    public static void main(String[] args) throws IOException {        new Main().run();    }    static final int INF = 0x3f3f3f3f;    int N, M;    double width(int[] X, int[] Y, int n, double x) {        double lb = INF, ub = -INF;        for (int i = 0; i < n; ++i) {            double x1 = X[i], y1 = Y[i], x2 = X[(i + 1) % n], y2 = Y[(i + 1) % n];            if ((x1 - x) * (x2 - x) <= 0 && x1 != x2) {                double y = y1 + (y2 - y1) * (x - x1) / (x2 - x1);                lb = Math.min(lb, y);                ub = Math.max(ub, y);            }        }        return Math.max(0, ub - lb);    }    int min(int[] nums, int start, int end) {        int min = INF;        for (int i = start; i < end; ++i) {            min = Math.min(min, nums[i]);        }        return min;    }    int max(int[] nums, int start, int end) {        int max = -INF;        for (int i = start; i < end; ++i) {            max = Math.max(max, nums[i]);        }        return max;    }    void solve(int[] X1, int[] Y1, int[] X2, int[] Y2) {        int min1 = min(X1, 0, N), max1 = max(X1, 0, N);        int min2 = min(X2, 0, M), max2 = max(X2, 0, M);        List<Integer> xs = new ArrayList<>();        for (int i = 0; i < N; ++i) xs.add(X1[i]);        for (int i = 0; i < M; ++i) xs.add(X2[i]);        Collections.sort(xs);        double res = 0.0;        for (int i = 0; i + 1 < xs.size(); ++i) {            double a = xs.get(i), b = xs.get(i + 1), c = (a + b) / 2;            if (min1 <= c && c <= max1 && min2 <= c && c <= max2) {                double fa = width(X1, Y1, N, a) * width(X2, Y2, M, a);                double fb = width(X1, Y1, N, b) * width(X2, Y2, M, b);                double fc = width(X1, Y1, N, c) * width(X2, Y2, M, c);                res += (b - a) / 6 * (fa + 4 * fc + fb);            }        }        out.printf("%.10f\n", res);    }    void read() {        while (true) {            N = ni();            M = ni();            if (N + M == 0) break;            int[] X1 = new int[N];            int[] Y1 = new int[N];            for (int i = 0; i < N; ++i) {                X1[i] = ni();                Y1[i] = ni();            }            int[] X2 = new int[M];            int[] Y2 = new int[M];            for (int i = 0; i < M; ++i) {                X2[i] = ni();                Y2[i] = ni();            }            solve(X1, Y1, X2, Y2);        }    }    FastScanner in;    PrintWriter out;    void run() throws IOException {        boolean oj;        try {            oj = ! System.getProperty("user.dir").equals("F:\\oxygen_workspace\\Algorithm");        } catch (Exception e) {            oj = System.getProperty("ONLINE_JUDGE") != null;        }        InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));        in = new FastScanner(is);        out = new PrintWriter(System.out);        long s = System.currentTimeMillis();        read();        out.flush();        if (!oj){            System.out.println("[" + (System.currentTimeMillis() - s) + "ms]");        }    }    public boolean more(){        return in.hasNext();    }    public int ni(){        return in.nextInt();    }    public long nl(){        return in.nextLong();    }    public double nd(){        return in.nextDouble();    }    public String ns(){        return in.nextString();    }    public char nc(){        return in.nextChar();    }    class FastScanner {        BufferedReader br;        StringTokenizer st;        boolean hasNext;        public FastScanner(InputStream is) throws IOException {            br = new BufferedReader(new InputStreamReader(is));            hasNext = true;        }        public String nextToken() {            while (st == null || !st.hasMoreTokens()) {                try {                    st = new StringTokenizer(br.readLine());                } catch (Exception e) {                    hasNext = false;                    return "##";                }            }            return st.nextToken();        }        String next = null;        public boolean hasNext(){            next = nextToken();            return hasNext;        }        public int nextInt() {            if (next == null){                hasNext();            }            String more = next;            next = null;            return Integer.parseInt(more);        }        public long nextLong() {            if (next == null){                hasNext();            }            String more = next;            next = null;            return Long.parseLong(more);        }        public double nextDouble() {            if (next == null){                hasNext();            }            String more = next;            next = null;            return Double.parseDouble(more);        }        public String nextString(){            if (next == null){                hasNext();            }            String more = next;            next = null;            return more;        }        public char nextChar(){            if (next == null){                hasNext();            }            String more = next;            next = null;            return more.charAt(0);        }    }}

注意width中for的循环，因为给定的坐标点已经是凸包形式，所以这种O(n)的更新方案才正确。