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
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：
代码如下：
代码语言：javascript复制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("%.10fn", 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。
