111 - DP LCS - degree1

2.解題思路：利用動態規劃的LCS演算法（演算法連結），特別注意要看清題目，第一行輸入為事件“envent”的數目，第二行為正確的事件"envent"的順序，第三行為學生的輸入

3.輸入輸出：sample的input、output就可以

4.程式碼

package HistoryGrading_111; import java.util.Scanner; public class Main { public static int length; public static int[] values; public static int[] paper; public static int[][] memo; public static void main(String[] args){ Scanner in = new Scanner(System.in); length = in.nextInt(); values = new int[length+1]; paper = new int[length+1]; for (int i = 1;i<=length;i++){ values[in.nextInt()] = i; } while (in.hasNext()){ for (int i=1;i<=length;i++){ paper[in.nextInt()] = i; } dp(); } } public static void dp(){ memo = new int[length+1][length+1]; int [][] m = memo; for (int i=0;i<=length;i++){ memo[i][0] = 0; memo[0][i] = 0; } for (int i=1;i<=length;i++){ for (int j=1;j<=length;j++){ if (paper[i] == values[j]){ memo[i][j] = memo[i-1][j-1] + 1; }else{ memo[i][j] = memo[i-1][j]>memo[i][j-1]?memo[i-1][j]:memo[i][j-1]; } } } System.out.println(memo[length][length]); } }

674 - DP 硬幣- degree1

2.解題思路：動態規劃的硬幣問題（連結）。

3.輸入輸出：sample的input、output就可以

4.程式碼：

package CoinChange_674; import java.io.IOException; import java.util.Scanner; public class Main { public static int[] changes = new int[]{1,5,10,25,50}; public static int des; public static int[] memo = new int[7490]; public static void main(String[] args) { memo[0] = 1; for (int i=0;i<=4;i++){ for (int j=changes[i];j<=7489;j++){ memo[j] = memo[j] + memo[j-changes[i]]; } } Scanner in = new Scanner (System.in); while (in.hasNext()){ des = in.nextInt(); System.out.println(memo[des]); } } }

562 - DP 非典型揹包- degree2

2.解題思路：這道題是一道非典型的DP題，我們不能直接利用DP得出答案，而是利用DP產生一箇中間的結果，然後利用這個中間結果計算題目答案。

  我們用DP思想首先產生一個judge陣列，陣列有0與1兩個值，judge[i] = 0表示我們無法劃分出i這個值，judge[i] = 1則相反表示利用一直的硬幣可以劃分出這個值。得到了judge陣列後我們在對陣列中所有值為1的數計算兩堆硬幣的差值，記錄下這個差值，輸出。

3.輸入輸出：sample

4.程式碼：

package DividingCoins_562; import java.util.Scanner; public class Main { public static int caseNum; public static int coinNum; public static int sum; public static int[] coins; public static int[] judge ; public static void main(String[] args) { Scanner in = new Scanner(System.in); caseNum = in.nextInt(); for(int i = 0;i<caseNum;i++){ sum = 0; coinNum = in.nextInt(); coins = new int[coinNum]; judge = new int[50000+1]; judge[0] = 1; for (int j=0;j<coinNum;j++){ coins[j] = in.nextInt(); sum = sum + coins[j]; for (int k = 50000-coins[j];k>=0;k--){ if (judge[k]==1){ judge[k+coins[j]] = 1; } } } int min = Integer.MAX_VALUE; for(int j = 0;j<=sum;j++){ if (judge[j]==1&&java.lang.Math.abs(2*j-sum)<min){ min = java.lang.Math.abs(2*j-sum); } } System.out.println(min); } } }

11137 - DP 硬幣 - degree-1

2.解題思路：典型動態規劃硬幣問題（連結），要特別注意迴圈接硬幣問題時的順序～見動態規劃硬幣演算法～

3.輸入輸出：sample

4.程式碼

package IngenuousCubrency_11137; import java.util.Scanner; public class Main { public static int[] value = new int[22]; public static long[] memo = new long[10001]; public static void main(String[] args) { for (int i=1;i<22;i++){ value[i] = i*i*i; } Scanner in = new Scanner(System.in); dp(); while (in.hasNextInt()){ System.out.println(memo[in.nextInt()]); } } public static void dp (){ memo[0] = 1; for (int j = 1 ;j<=10000;j++){ for (int i=1;i<=21;i++){ if(j>=value[i]){ memo[j] = memo[j] + memo[j -value[i]]; } } } } }  

10534 - DP LIS - degree -2

2.解題思路：這道題主要是利用了LIS演算法（LIS演算法連結），要特別注意o(n^2)與o(nlogn)兩種演算法的不同實現，這道題用o(n^2)的方法

是超時的，只能使用o(nlogn)的方法。、

  演算法的思路是找到目標序列的中點，以這個中點為起點與中點的LIS是一樣的。我們從input序列的中點每次分別向左與向右同時擴充套件一個點，計算LIS並且比較，如果相等並且超過以後點能生成的最大值則為答案。

3.輸入輸出：sample

4.程式碼:

package WavioSequence_10534; import java.util.Scanner; public class Main { static int length; static int[] memo; static int[] memoReverse; static int[] sequence; static int[] sequenceReverse ; static int[] min; static int[] minReverse; public static void main(String[] args) { Scanner in = new Scanner(System.in); while (in.hasNext()){ length = in.nextInt(); sequence = new int[length+1]; sequenceReverse = new int[length+1]; for(int i=1;i<=length;i++){ sequence[i] = in.nextInt(); sequenceReverse[length+1-i] = sequence[i]; } dpQuick(); int max = max(); System.out.println(2*max-1); } } public static void dpQuick(){ min = new int[length+1]; minReverse = new int[length+1]; memo = new int[length+1]; memoReverse = new int[length+1]; min[0] = -1; minReverse[0] = -1; int len = 0; int lenReverse = 0; for (int i = 1;i<=length;i++){ if (sequence[i]>min[len]){ min[++len] = sequence[i]; memo[i] = len; }else{ int j = binarySearch(min,1,len-1,sequence[i]); min[j] = sequence[i]; memo[i] = j; } if (sequenceReverse[i]>minReverse[lenReverse]){ minReverse[++lenReverse] = sequenceReverse[i]; memoReverse[i] = lenReverse; }else{ int j = binarySearch(minReverse,1,lenReverse-1,sequenceReverse[i]); minReverse[j] = sequenceReverse[i]; memoReverse[i] = j; } } } public static int binarySearch(int[] min,int begin,int end,int value){ if (begin<=end){ int mid = begin + (end - begin)/2; if (min[mid] == value){ return mid; }else if(min[mid]>value){ end = mid -1; return binarySearch(min,begin,end,value); }else{ begin = mid + 1; return binarySearch(min,begin,end,value); } }else{ return begin; } } public static int max(){ int max = 0; boolean stop = false; for (int offset=(1+length)/2-1;offset>=0&&!stop;offset--){ int left = 1+offset; int right = length - offset; int rightLIS = memo[right]<memoReverse[left]?memo[right]:memoReverse[left]; int leftLIS = memo[left]<memoReverse[right]?memo[left]:memoReverse[right]; max = max>rightLIS?max:rightLIS; max = max>leftLIS?max:leftLIS; if(max>=offset){ stop = true; } } return max; } }  

10130 - DP 0-1揹包問題 - degree -2

2.解題思路：這是一道典型的0-1揹包問題，不過對於家庭裡的每個成員都需要計算一次，然後將每個成員計算出的最大值累加輸出。

3.輸入輸出：sample

4.程式碼:

package SuperSale_10130; import java.util.Scanner; import com.sun.org.apache.xerces.internal.impl.dv.ValidatedInfo; public class Main { /** * @param args */ public static int[] price; public static int[] weight; public static int[] hold; public static int[][] memo; public static int kind; public static int member; public static int max = 0; public static void main(String[] args) { Scanner in = new Scanner (System.in); int loopTimes = in.nextInt(); for (int i=1;i<=loopTimes;i++){ kind = in.nextInt(); price = new int[kind+1]; weight = new int[kind+1]; for (int j=1;j<=kind;j++){ price[j] = in.nextInt(); weight[j] = in.nextInt(); } member = in.nextInt(); hold = new int[member+1]; for (int j=1;j<=member;j++){ hold[j] = in.nextInt(); if(hold[j]>max){ max = hold[j]; } } memo = new int[max+1][kind+1]; dp(); System.out.println(dp()); } } public static int dp(){ int [][] tempmemo = memo; int [] tempweight = weight; int [] tempprice = price; for (int k=kind;k>=1;k--){ for (int j = 0;j<=max;j++){ if (k == kind){ if (j>=weight[k]) memo[j][k] = price[k]; }else{ memo[j][k] = memo[j][k+1]; if (j>=weight[k]) memo[j][k] = memo[j][k]>memo[j-weight[k]][k+1]+price[k]?memo[j][k]:memo[j-weight[k]][k+1]+price[k]; } } } int sum = 0; for (int i = 1;i<=member;i++){ int max =0; for(int j=1;j<=kind;j++){ if (memo[hold[i]][j]>max){ max = memo[hold[i]][j]; } } sum = sum + max; } return sum; } }