挑战程序竞赛系列（31）：4.5剪枝
来源：互联网 发布：旅行收纳袋 知乎 编辑：程序博客网 时间：2024/05/22 04:40
挑战程序竞赛系列（31）：4.5剪枝
详细代码可以fork下Github上leetcode项目，不定期更新。
练习题如下：
- POJ 1011: Sticks
- POJ 2046: Gap
- POJ 3134: Power Calculus
POJ 1011: Sticks
变态的DFS搜索，需要剪枝否则TLE，初始版本如下：
void solve() { while (true){ int n = ni(); if (n == 0) break; int[] sticks = new int[n]; int min = 0; int sum = 0; int max = 0; for (int i = 0; i < n; ++i){ int len = ni(); max = Math.max(max, len); sum += len; sticks[i] = len; } Arrays.sort(sticks); Set<Integer> mem = new HashSet<Integer>(); for (int i = n; i >= 1; --i){ if (sum % i == 0){ min = sum / i; if (min >= max && dfs(sticks, min, 0, new boolean[n], mem)) break; } } out.println(min); } } public boolean dfs(int[] sticks, int min, int sum, boolean[] visited, Set<Integer> mem){ if (mem.size() == sticks.length){ return min == sum; } if (sum > min) return false; if (sum == min){ if (dfs(sticks, min, 0, visited, mem)) return true; else return false; } for (int i = sticks.length - 1; i >= 0; --i){ int rem = min - sum; if (!visited[i] && sticks[i] <= rem){ visited[i] = true; mem.add(i); if (dfs(sticks, min, sum + sticks[i], visited, mem)){ return true; } else{ visited[i] = false; mem.remove(i); } } } return false; }代码细节可以忽略，visited和mem可以合并，做了一些简单的剪枝处理，但始终超时。思路是遍历各种组合，且当所有元素被使用后，看是否能够找到所有长度一致的木棒。
遍历超时很大一部分的原因在于dfs中有个for循环，对于重复长度的棒子过滤的不够干净，浪费了大量的搜素资源。我们可以采用map对长度进行统计，这样重复长度的棒子大可不必搜索，省时省力。
依旧遍历，对每种可能的组合进行搜索，搜索时记录拼接完成的棒子个数，个数 * 可能长度 = 总和时，遍历结束。
具体细节可以参考博文：http://www.hankcs.com/program/algorithm/poj-1011-sticks.html，不作赘述。
代码如下：
import java.io.File;import java.io.FileInputStream;import java.io.IOException;import java.io.InputStream;import java.io.PrintWriter;import java.util.Arrays;import java.util.HashSet;import java.util.InputMismatchException;import java.util.Set;public class Main{ InputStream is; PrintWriter out; String INPUT = "./data/judge/201707/P1011.txt"; int[] in; int candicate; void solve() { while (true){ int n = ni(); if (n == 0) break; in = new int[51]; finish = false; int sum = 0; candicate = 0; int max = 0; for (int i = 0; i < n; ++i){ int len = ni(); max = Math.max(max, len); sum += len; in[len] ++; } candicate = max; while (true){ if (sum % candicate == 0){ check(sum / candicate, candicate, max); } if (finish) break; ++candicate; } out.println(candicate); } } boolean finish; public void check(int count, int len, int plen){ --in[plen]; if (count == 0){ finish = true; } if (!finish){ len -= plen; //剩余长度 if (len != 0){ int nextPlen = Math.min(len, plen); for (; nextPlen > 0; --nextPlen){ if (in[nextPlen] != 0){ check(count, len, nextPlen); } } } else{ int max = 50; while (max > 0 && in[max] == 0) --max; check(count - 1, candicate, max); //当前剩余棒子的最大长度 } } ++in[plen]; } void run() throws Exception { is = oj ? System.in : new FileInputStream(new File(INPUT)); out = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); out.flush(); tr(System.currentTimeMillis() - s + "ms"); } public static void main(String[] args) throws Exception { new Main().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte() { if (lenbuf == -1) throw new InputMismatchException(); if (ptrbuf >= lenbuf) { ptrbuf = 0; try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if (lenbuf <= 0) return -1; } return inbuf[ptrbuf++]; } private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while ((b = readByte()) != -1 && isSpaceChar(b)) ; return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char) skip(); } private String ns() { int b = skip(); StringBuilder sb = new StringBuilder(); while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' // ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n) { char[] buf = new char[n]; int b = skip(), p = 0; while (p < n && !(isSpaceChar(b))) { buf[p++] = (char) b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m) { char[][] map = new char[n][]; for (int i = 0; i < n; i++) map[i] = ns(m); return map; } private int[] na(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = ni(); return a; } private int ni() { int num = 0, b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private long nl() { long num = 0; int b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private boolean oj = System.getProperty("ONLINE_JUDGE") != null; private void tr(Object... o) { if (!oj) System.out.println(Arrays.deepToString(o)); }}提供一组变态数据，可以自己测测来玩玩，上述算法需要3s搜索出答案。
6440 40 30 35 35 26 15 40 40 40 40 40 40 40 40 40 40 40 40 40 4040 40 43 42 42 41 10 4 40 40 40 40 40 40 40 40 40 40 40 40 4040 25 39 46 40 10 4 40 40 37 18 17 16 15 40 40 40 40 40 40 40 400ans:454[3503ms]POJ 只需128ms走完全部测试数据，数据有点水啊。
POJ 2046: Gap
一道模拟题，用BFS广搜就好了，关键抓住填入空格的规则，只有一种情况，只允许填入左侧的下一个数字，所以在当前board下只会出现四种状态，没有什么搜索策略，按照轮次搜即可。
BFS的一个好处在于，能够以最短的距离搜到终止状态，也是此题的关键。不过还需要注意，当我们定义board的状态时，可以从整体出发，需要重写hashCode和equal方法，方便记录状态的访问情况，好题。
步骤：
- 定义状态
- 考虑状态的终止条件
- 考虑状态的切换规则
- 重写hashCode和equals方法
代码如下：
import java.io.File;import java.io.FileInputStream;import java.io.IOException;import java.io.InputStream;import java.io.PrintWriter;import java.util.Arrays;import java.util.HashSet;import java.util.InputMismatchException;import java.util.LinkedList;import java.util.Queue;import java.util.Set;public class Main{ InputStream is; PrintWriter out; String INPUT = "./data/judge/201708/P2046.txt"; class Game{ int[][] board = new int[4][8]; int turn; public Game(int[][] board){ this.board = board; this.turn = 0; int[] y = find(11); swap(new int[]{0, 0}, y); y = find(21); swap(new int[]{1, 0}, y); y = find(31); swap(new int[]{2, 0}, y); y = find(41); swap(new int[]{3, 0}, y); } public Game(Game newGame){ for (int i = 0; i < 4; ++i){ for (int j = 0; j < 8; ++j){ this.board[i][j] = newGame.board[i][j]; } } this.turn = newGame.turn; } public boolean canFill(int i, int j){ if (board[i][j] != 0) return false; if (board[i][j - 1] != 0 && (board[i][j - 1] % 10) != 7) return true; return false; } public boolean done(){ for (int i = 0; i < 4; ++i){ if (board[i][7] != 0) return false; } for (int i = 0; i < 4; ++i){ for (int j = 0; j < 7; ++j){ if (board[i][j] != (i + 1) * 10 + (j + 1)) return false; } } return true; } public void fillGap(int i, int j){ int key = board[i][j - 1] + 1; int[] pos = find(key); swap(new int[]{i, j}, pos); this.turn ++; } @Override public boolean equals(Object obj) { if (obj instanceof Game){ Game that = (Game)obj; for (int i = 0; i < 4; ++i){ for (int j = 0; j < 8; ++j){ if (board[i][j] != that.board[i][j]) return false; } } return true; } else return false; } public int[] find(int key){ for (int i = 0; i < 4; ++i){ for (int j = 0; j < 8; ++j){ if (board[i][j] == key) return new int[]{i, j}; } } return new int[]{-1, -1}; } public void swap(int[] x, int[] y){ int tmp = board[x[0]][x[1]]; board[x[0]][x[1]] = board[y[0]][y[1]]; board[y[0]][y[1]] = tmp; } @Override public String toString() { StringBuilder sb = new StringBuilder(); for (int i = 0; i < 4; ++i){ for (int j = 0; j < 8; ++j){ sb.append(board[i][j] + (j + 1 == 8 ? "\n" : " ")); } } return sb.toString(); } @Override public int hashCode() { int hash = 0; for (int i = 0; i < 4; ++i){ for (int j = 1; j < 8; ++j){ hash += board[i][j]; hash <<= 1; } } return hash; } } void solve() { int T = ni(); while (T --> 0){ int[][] board = new int[4][8]; for (int i = 0; i < 4; ++i){ for (int j = 1; j < 8; ++j){ board[i][j] = ni(); } } Game game = new Game(board); Queue<Game> queue = new LinkedList<Game>(); Set<Game> visited = new HashSet<Game>(); if (game.done()){ out.println(0); continue; } queue.offer(game); int ans = -1; boolean end = false; outer: while (!queue.isEmpty() && !end){ Game gg = queue.poll(); if (visited.contains(gg)) continue; visited.add(gg); for (int i = 0; i < 4; ++i){ for (int j = 1; j < 8; ++j){ if (gg.canFill(i, j)){ Game tmp = new Game(gg); tmp.fillGap(i, j); if (tmp.done()){ ans = tmp.turn; end = true; continue outer; } else queue.offer(tmp); } } } } out.println(ans); } } void run() throws Exception { is = oj ? System.in : new FileInputStream(new File(INPUT)); out = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); out.flush(); tr(System.currentTimeMillis() - s + "ms"); } public static void main(String[] args) throws Exception { new Main().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte() { if (lenbuf == -1) throw new InputMismatchException(); if (ptrbuf >= lenbuf) { ptrbuf = 0; try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if (lenbuf <= 0) return -1; } return inbuf[ptrbuf++]; } private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while ((b = readByte()) != -1 && isSpaceChar(b)) ; return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char) skip(); } private String ns() { int b = skip(); StringBuilder sb = new StringBuilder(); while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' // ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n) { char[] buf = new char[n]; int b = skip(), p = 0; while (p < n && !(isSpaceChar(b))) { buf[p++] = (char) b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m) { char[][] map = new char[n][]; for (int i = 0; i < n; i++) map[i] = ns(m); return map; } private int[] na(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = ni(); return a; } private int ni() { int num = 0, b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private long nl() { long num = 0; int b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private boolean oj = System.getProperty("ONLINE_JUDGE") != null; private void tr(Object... o) { if (!oj) System.out.println(Arrays.deepToString(o)); }}POJ 3134: Power Calculus
第一次遇到迭代加深算法，有点难以理解。刚开始使用BFS，但发现细节处理上有问题。此题有些关键地方，比如同一轮生成的解，不能结合使用，只能使用前几层的解和当前层解的组合，或许可以如此想象，在构造轮次时，只有一条链，这种构造路径难道不是DFS？没错，就是它，但是何时终止呢？
剪枝算法告诉我们，每个给定的n都有一个上界，就拿快速幂的例子来说，举例13，至多也就这些操作：
13-1=1212/2=66/2=33-1=22/2=1于是我们迭代找解的时候可以根据此上界进行剪枝，对数据预处理下，有上界函数：
public int upper(int n){ int cnt = 0; while (n > 0){ if ((n & 1) != 0){ cnt ++; } n >>= 1; cnt ++; } return cnt - 2; }接着就是构一条生成路径了，采用DFS，巧妙之处在于这种DFS刚好能够模拟这种构造状态，神奇，比如：
1 -> 0表示x需要耗费0次，初始状态那么自然地：2 = 1 + 1表示： x^2 = x * x所以有1生成了2那么3怎么来？ 1 + 2 = 3但是当前层还会有 2 + 2 = 4但神奇的是，1 + 2 = 3 和 2 + 2 = 4不会同一时刻遍历，而是分为两次dfs调用。这就解决了之前BFS的一个bug，呵呵哒。代码如下：
import java.io.File;import java.io.FileInputStream;import java.io.IOException;import java.io.InputStream;import java.io.PrintWriter;import java.util.Arrays;import java.util.InputMismatchException;public class Main{ InputStream is; PrintWriter out; String INPUT = "./data/judge/201708/P3134.txt"; int MAX_N = 1024; int MAX_D = 20; int[] exp = new int[MAX_D]; int[] ans = new int[MAX_N]; void solve(){ Arrays.fill(exp, 1); for (int i = 2; i < MAX_N; ++i){ ans[i] = upper(i); } dfs(0); while (true){ int n = ni(); if (n == 0) break; out.println(ans[n]); } } public void dfs(int d) { if (d > MAX_D) { return; } for (int i = 0; i <= d; i++) { exp[d + 1] = exp[i] + exp[d]; // 乘法 if (exp[d + 1] < MAX_N && ans[exp[d + 1]] >= d + 1) { //这层的解要是被更新的话，继续更新下下层的解 ans[exp[d + 1]] = d + 1; //更新解 dfs(d + 1); } exp[d + 1] = exp[d] - exp[i]; // 除法 if (exp[d + 1] > 0 && ans[exp[d + 1]] >= d + 1) { ans[exp[d + 1]] = d + 1; dfs(d + 1); } } } public int upper(int n){ int cnt = 0; while (n > 0){ if ((n & 1) != 0){ cnt ++; } n >>= 1; cnt ++; } return cnt - 2; } void run() throws Exception { is = oj ? System.in : new FileInputStream(new File(INPUT)); out = new PrintWriter(System.out); long s = System.currentTimeMillis(); solve(); out.flush(); tr(System.currentTimeMillis() - s + "ms"); } public static void main(String[] args) throws Exception { new Main().run(); } private byte[] inbuf = new byte[1024]; public int lenbuf = 0, ptrbuf = 0; private int readByte() { if (lenbuf == -1) throw new InputMismatchException(); if (ptrbuf >= lenbuf) { ptrbuf = 0; try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if (lenbuf <= 0) return -1; } return inbuf[ptrbuf++]; } private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); } private int skip() { int b; while ((b = readByte()) != -1 && isSpaceChar(b)) ; return b; } private double nd() { return Double.parseDouble(ns()); } private char nc() { return (char) skip(); } private String ns() { int b = skip(); StringBuilder sb = new StringBuilder(); while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' // ') sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private char[] ns(int n) { char[] buf = new char[n]; int b = skip(), p = 0; while (p < n && !(isSpaceChar(b))) { buf[p++] = (char) b; b = readByte(); } return n == p ? buf : Arrays.copyOf(buf, p); } private char[][] nm(int n, int m) { char[][] map = new char[n][]; for (int i = 0; i < n; i++) map[i] = ns(m); return map; } private int[] na(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = ni(); return a; } private int ni() { int num = 0, b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private long nl() { long num = 0; int b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } private boolean oj = System.getProperty("ONLINE_JUDGE") != null; private void tr(Object... o) { if (!oj) System.out.println(Arrays.deepToString(o)); }}这就厉害了，类似于BFS，但能够精确的控制每层解的非法组合。
阅读全文
0 0
- 挑战程序竞赛系列（31）：4.5剪枝
- 挑战程序竞赛系列（32）：4.5 A*与IDA*
- 挑战程序竞赛系列（81）：4.3 LCA（1）
- 挑战程序竞赛系列（82）：4.3 LCA（2）
- 挑战程序竞赛系列（1）：2.3动态规划
- 挑战程序竞赛系列（4）：2.1深度优先搜索
- 挑战程序竞赛系列（5）：2.1广度优先搜索
- 挑战程序竞赛系列（6）：2.1穷尽搜索
- 挑战程序竞赛系列（9）：2.4优先队列
- 挑战程序竞赛系列（10）：2.4并查集
- 挑战程序竞赛系列（11）：2.5最短路径
- 挑战程序竞赛系列（12）：2.5最小生成树
- 挑战程序竞赛系列（13）：2.6辗转相除法
- 挑战程序竞赛系列（14）：2.6素数
- 挑战程序竞赛系列（15）：2.6快速幂运算
- 挑战程序竞赛系列（16）：3.1最大化最小值
- 挑战程序竞赛系列（17）：3.1最大化平均值
- 挑战程序竞赛系列（20）：3.2尺取法
- HDU 4185 Oil Skimming 二分匹配
- Hdu 6060 RXD and dividing【思维】
- 冒泡排序和插入排序
- 【转】快速搭建react的webpack初始环境
- Javascript整理
- 挑战程序竞赛系列（31）：4.5剪枝
- var 有无的区别
- ccf认证窗口
- openvpn配置
- LeetCode oj 21. Merge Two Sorted Lists（分类：链表）(难度：easy)
- Unity教程之-Unity3d自带寻路Navmesh
- 搜索专题总结
- shell 脚本编程之算术运算符
- 关于MFC当中使用VC(C++)8位BMP提取Y通道平均值的笔记
