USACO Section1.2

section1.1主要包括四道题和两个编程知识介绍。下面将对这6个部分内容进行学习。

Your Ride Is Here

这道题没什么难度，读懂题目意思就行：把两个字符串按照题目要求转换成数字，然后对47取模，比较是否相等。
代码如下

/*
LANG: JAVA
TASK: ride
*/

import java.io.*;
import java.util.*;
public class ride {
    public static void main (String [] args) throws IOException {
        BufferedReader f = new BufferedReader(new FileReader("ride.in"));
        PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("ride.out")));

        Scanner in = new Scanner(f);
        String comet = null;
        String group = null;
        while (in.hasNextLine()) {
            comet = in.nextLine();
            group = in.nextLine();
        }
        if (caculate(group) == caculate(comet)) {
            out.println("GO");
        }else {
            out.println("STAY");
        }
        out.close();                                  // close the output file
    }

    public static int caculate(String str) {
        int size = str.length();
        int result = 1;
        for (int i = 0; i < size; i++) {
            result = result * (str.charAt(i) - 'A' + 1);
        }
        return result % 47;
    }
}

Programming Contest Problem Types

这个内容介绍了编程比赛中的题目类型，Hal Burch有研究发现：编程比赛里面只有16种比赛类型。这16种比赛类型的前几个几乎占了IOI比赛题目的80%。
这16种题型分别是：

• Dynamic Programming
• Greedy
• Complete Search
• Flood Fill
• Shortest Path
• Recursive Search Techniques
• Minimum Spanning Tree
• Knapsack
• Computational Geometry
• Network Flow
• Eulerian Path
• Two-Dimensional Convex Hull
• BigNums
• Heuristic Search
• Approximate Search
• Ad Hoc Problems

稍微复杂一点的程序就结合例上述多个类型，所以呀，想把USACO刷完，还是挺不容易的。

Ad Hoc Problems

这种问题也叫“临时问题”，是上面16种类型中最后一种。它不属于标准的，具有精心研究的解决方案，反而是每个问题都没有特定的方法来解决。

这种问题一般需要仔细阅读，也需要一些合理的优化和一些分析。解决方案可能需要一个新颖的数据结构等。

Greedy Gift Givers

这道题就是典型的Ad Hoc Problems问题。题目的大概意思是分礼物。难点在于读懂题目和如何读数据。难度倒是不大，代码如下

/*
LANG: JAVA
TASK: gift1
*/

import java.io.*;
import java.util.*;
public class gift1 {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new FileReader("gift1.in"));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("gift1.out")));
        int size = Integer.parseInt(br.readLine());
        String[] names = new String[size];
        HashMap<String, Integer> net = new HashMap<String, Integer>();
        for (int i = 0; i < size; i++) {
            String name = br.readLine();
            net.put(name, 0);
            names[i] = name;
        }
        for (int i = 0; i < size; i++) {
            String name = br.readLine();
            String gift = br.readLine();
            String[] temp = gift.split(" ");
            int sum = Integer.parseInt(temp[0]);
            int div = Integer.parseInt(temp[1]);
            if (div != 0) {
                for (int j = 0; j < div; j++) {
                    String target = br.readLine();
                    net.put(target, net.get(target) + sum / div);
                }
                net.put(name, net.get(name) - (sum - sum % div));
            }
        }
        for (String nameToPrint : names) {
            pw.println(nameToPrint + " " + net.get(nameToPrint));
        }
        pw.close();
        br.close();
    }

}

Friday the Thirteenth

这道题也是需要读懂题目。判断闰年和计算天数的函数比较简单，关键是最后输出的时候，数组里面的值分别对应星期几容易出错。

/*
LANG: JAVA
TASK: friday
*/

import java.io.*;
public class friday {

    public static void main(String[] args) throws Exception {
        int count[] = new int[7];
        BufferedReader br = new BufferedReader(new FileReader("friday.in"));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("friday.out")));
        int years = Integer.parseInt(br.readLine());
        int allDays = 0;
        for (int i = 1900; i < 1900 + years; i++) {
            for (int j = 1; j <= 12; j++) {
                count[(allDays + 13) % 7]++;
                allDays += countDaysInMonth(j, i);
            }
        }
        pw.println(count[6] + " " + count[0] + " " + count[1] + " " + count[2] + " " + count[3] + " " + count[4] + " " + count[5]);
        pw.close();
        br.close();
    }
    public static boolean isLeapYear(int year) {
        if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0)) {
            return true;
        }else {
            return false;
        }
    }
    public static int countDaysInMonth(int month, int year) {
        switch (month) {
            case 4:
            case 6:
            case 9:
            case 11:
                return 30;
            case 2:
                if (isLeapYear(year)) {
                    return 29;
                }else {
                    return 28;
                }
            default:
                return 31;
        }
    }
}

Broken Necklace

这个题算的上是secion1.1里最难的题了。题目也不太好读懂。看懂题目后直观感觉是暴力求解。这是我写的代码。

/*
LANG: JAVA
TASK: beads
*/

import java.io.*;
public class beads {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new FileReader("beads.in"));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("beads.out")));
        int numOfBeads = Integer.parseInt(br.readLine());
        String necklace = br.readLine();
        necklace = necklace + necklace;
        int maxTotal = 0;
        for (int i = 1; i < necklace.length() - 1; i++) {
            int total = 0;
            char start = necklace.charAt(i);
            char end = necklace.charAt(i - 1);
            //找到有意义的起始点
            for (int j = i; j < necklace.length() - 1; j++) {
                if (necklace.charAt(j) == 'w') {
                    continue;
                } else {
                    start = necklace.charAt(j);
                    break;
                }
            }

            for (int j = i; j < necklace.length() - 1; j++) {
                if (necklace.charAt(j) == 'w') {
                    total++;
                } else {
                    if (necklace.charAt(j) == start) {
                        total++;
                    }else {
                        break;
                    }
                }
            }
            //找到有意义的起始点
            for (int j = i - 1; j >= 0; j--) {
                if (necklace.charAt(j) == 'w') {

                    continue;
                } else {
                    end = necklace.charAt(j);
                    break;
                }
            }

            for (int j = i - 1; j >= 0; j--) {
                if (necklace.charAt(j) == 'w') {
                    total++;
                } else {
                    if (necklace.charAt(j) == end) {
                        total++;
                    }else {
                        break;
                    }
                }
            }

            maxTotal = maxTotal > total?maxTotal:total;

        }
        maxTotal = maxTotal < numOfBeads?maxTotal:numOfBeads;
        pw.println(maxTotal);

        pw.close();
        br.close();
    }

}

如果要算复杂度的话，估计有4×N^2，可能比一般人写的更慢。提交通过后看了官网给的答案，有一个复杂度为O(n)的方法。

/*
LANG: JAVA
TASK: beads
*/

import java.io.*;
public class beads1 {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new FileReader("beads.in"));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("beads.out")));
        int numOfBeads = Integer.parseInt(br.readLine());
        String necklace = br.readLine();

        necklace = necklace + necklace;
        int[][] left = new int[necklace.length() + 1][2];
        int[][] right = new int[necklace.length() + 1][2];
        left[0][0] = left[0][1] = 0;

        for (int i = 1; i <= necklace.length(); i++) {
            if (necklace.charAt(i - 1) == 'r') {
                left[i][0] = left[i - 1][0] + 1;//第0列存r，假如为r，则加一
                left[i][1] = 0;//第1列存b，假如为r，则置为0
            } else if (necklace.charAt(i - 1) == 'b') {
                left[i][1] = left[i - 1][1] + 1;
                left[i][0] = 0;
            } else {
                left[i][0] = left[i - 1][0] + 1;
                left[i][1] = left[i - 1][1] + 1;
            }
        }

        right[necklace.length()][0] = right[necklace.length()][1] = 0;
        for (int i=necklace.length() - 1; i >= 0; i--){
            if (necklace.charAt(i) == 'r'){
                right[i][0] = right[i + 1][0] + 1;//第0列存r，假如为r，则加一
                right[i][1] = 0;//第1列存b，假如为r，则置为0
            } else if (necklace.charAt(i) == 'b'){
                right[i][1] = right[i + 1][1] + 1;
                right[i][0] = 0;
            } else {
                right[i][0] = right[i + 1][0] + 1;
                right[i][1] = right[i + 1][1] + 1;
            }
        }

        int m = 0;
        for (int i = 0; i < necklace.length(); i++) {
            m = Math.max(m, Math.max(left[i][0], left[i][1]) + Math.max(right[i][0], right[i][1]));
        }
        m = Math.min(m, numOfBeads);
        pw.println(m);
        pw.close();
        br.close();
    }

}