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详细代码可以fork下Github上leetcode项目,不定期更新。
练习题如下:
题意参考博文:
http://www.hankcs.com/program/algorithm/aoj-2214-hall-warp.html
思路:
hankcs讲的很清楚,先对虫洞的起始位置排序,接着用一个dpi表示到达第i个虫洞起始位置所需要的方案数。
首先,如果不考虑i前面的虫洞,它的方案数有C(x1+y1,x1)C(x1 + y1, x1),现在在i前面有i-1个虫洞,该如何组合呢?
可以分为两种情况:
dp[i] = dp[i] + dp[j] * (j终点到i起点的方案数 - j起点到i起点的方案数) j = 0, 1, ..., i - 1;
dp[i] = dp[i] + dp[j] * (0 - j起点到i起点的方案数) j = 0, 1, ..., i - 1;
因为从虫洞j不可能回到虫洞i,所以j终点到i起点的方案数为0
ok,有了递推式,现在只要求解cnk模,《挑战》P294有关于CNK取模的算法,但这针对的是p不一定是素数的情况,而此题MOD = 1000000007,是个大素数,所以可以直接取fact的逆,公式为:
CKN≡a1(a2a3)−1
C_N^K \equiv a_1(a_2a_3)^{-1}
代码如下:
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;
public class Main{
InputStream is;
PrintWriter out;
String INPUT = "./data/judge/201709/A2214.txt";
static Scanner in;
public static void main(String[] args) throws Exception {
in = new Scanner(System.in);
new Main().solve();
}
static final int MOD = 1000000007;
class Wormhole implements Comparable<Wormhole> {
int x1;
int y1;
int x2;
int y2;
public Wormhole(int x1, int y1, int x2, int y2) {
this.x1 = x1;
this.x2 = x2;
this.y1 = y1;
this.y2 = y2;
}
@Override
public int compareTo(Wormhole that) {
return this.x1 != that.x1 ? this.x1 - that.x1 : this.y1 - that.y1;
}
}
List<Wormhole> holes;
void solve() {
init_factorial();
while (true) {
int n = in.nextInt();
int m = in.nextInt();
int k = in.nextInt();
if (n == 0 && m == 0 && k == 0) break;
holes = new ArrayList<>();
n--;
m--;
for (int i = 0; i < k; ++i) {
int x1 = in.nextInt();
int y1 = in.nextInt();
int x2 = in.nextInt();
int y2 = in.nextInt();
holes.add(new Wormhole(x1 - 1, y1 - 1, x2 - 1, y2 - 1));
}
Collections.sort(holes);
holes.add(new Wormhole(n, m, n + 1, m + 1));
int[] dp = new int[k + 16];
for (int i = 0; i <= k; ++i) {
Wormhole hole = holes.get(i);
dp[i] = nck(hole.x1 + hole.y1, hole.x1);
for (int j = 0; j < i; ++j) {
Wormhole prev = holes.get(j);
dp[i] = add(dp[i], mul(dp[j],
sub(calc(prev.x2, prev.y2, hole.x1, hole.y1), calc(prev.x1, prev.y1, hole.x1, hole.y1))));
}
}
System.out.println(dp[k]);
}
}
/************************* mod_fact && mod_comb *****************************/
int add(int a, int b) {
return (a + b) % MOD;
}
int sub(int a, int b) {
return (a - b + MOD) % MOD;
}
int mul(long a, long b) {
return (int) (((a * b)) % MOD);
}
int[] fact = new int[200010];
void init_factorial() {
fact[0] = 1;
for (int i = 1; i < 200000; ++i) {
fact[i] = mul(fact[i - 1], i);
}
}
class GCD {
int d;
int x;
int y;
public GCD(int d, int x, int y) {
this.d = d;
this.x = x;
this.y = y;
}
}
GCD extgcd(int a, int b) {
if (b == 0)
return new GCD(a, 1, 0);
else {
GCD p = extgcd(b, a % b);
GCD ans = new GCD(0, 0, 0);
ans.d = p.d;
ans.x = p.y;
ans.y = p.x - (a / b) * p.y;
return ans;
}
}
int inv(int a, int m) {
GCD p = extgcd(a, m);
if (p.d != 1)
return -1;
return (p.x % m + m) % m;
}
int nck(int n, int k) {
return mul(mul(fact[n], inv(fact[n - k], MOD)), inv(fact[k], MOD));
}
int calc(int x1, int y1, int x2, int y2) {
if (x2 < x1 || y2 < y1)
return 0;
return nck(x2 - x1 + y2 - x1, x2 - x1);
}
}
测试用例都过了,但放在AOJ上就RE,总感觉测试接口有问题。。。