挑战程序竞赛系列（49）：4.2 推理与动态规划算法（2）

挑战程序竞赛系列（49）：4.2 推理与动态规划算法（2）

详细代码可以fork下Github上leetcode项目，不定期更新。

练习题如下：

POJ 2068: Nim

POJ 2068: Nim

团体尼姆赛：传统的尼姆游戏由两名玩家进行，在一堆石头中，双方轮流取走任意合法数量块石头，取走最后一块石头的玩家落败。多人尼姆游戏将参赛人数拓展至两个队伍，每支队伍有n名队员交错入座，单次分别能最多取走Mi块石头，取走S块石头中的最后一块的队伍失败，求第一支队伍是否有必胜策略？

动态规划题，如果单纯的两个人轮流取，直接利用数学公式直接求出，但此题多轮多人且每人取的石头数也不同，这貌似只能动规了。

dp[i][j][k] 第i支队伍第k个人，剩余k个石子时，能否赢得当前轮

注意当k = 0，表明是必胜态，而当k = 1时，一定为必输态。

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.StringTokenizer;

public class Main{

String INPUT = "./data/judge/201709/P2068.txt";

public static void main(String[] args) throws IOException {
new Main().run();
}

static final int MAX_S = 1 << 13;
boolean[][][] dp; //第k只队伍 第j个队员 剩余i个石头的状态

void solve() {
while (true) {
int n = ni();
if (n == 0) break;
int s = ni();

dp = new boolean[2][12][MAX_S + 16];

int[] team = new int[2 * n];
for (int i = 0; i < 2 * n; ++i) {
team[i] = ni();
}

for (int i = 0; i < 2 * n; ++i) {
dp[i & 1][i / 2 + 1][0] = true;
dp[i & 1][i / 2 + 1][1] = false; // 必输态
}

for (int i = 2; i <= s; ++i) {
for (int now = 0; now < 2 * n; ++now) {
int nxt = (now + 1) % (2 * n);
for (int j = team[now]; j >= 1; --j) {
if (i - j >= 0) {
dp[now & 1][now / 2 + 1][i] |= !dp[nxt & 1][nxt / 2 + 1][i - j];
if (dp[now & 1][now / 2 + 1][i]) break;
}
}
}
}

out.println(dp[0][1][s] ? "1" : "0");
}
}

FastScanner in;
PrintWriter out;

void run() throws IOException {
boolean oj;
try {
oj = ! System.getProperty("user.dir").equals("F:\\java_workspace\\leetcode");
} 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();
solve();
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);
}
}
}