摘要： E 1.1.1:Compilier: translate code from source language to target languageInterpreter: translate code + execution (code + input + interpreter -> output)E1.1.2:Pro of compiler: simpler design; multiple runPro of interpreter: realtime runningE1.1.3:Easier to debug the compilerCan do an extra level o 阅读全文

摘要： 34.1-1:Assumption: the graph is unweightedLONGEST-PATH-LENGTH ∈P=> LONGEST-PATH∈P:Suppose there's a problem solves LHS in polynomial time, gives result k; then for the problem of RHS, it returns true if the given k' <= k.LONGEST-PATH∈P:=> LONGEST-PATH-LENGTH ∈P:Guess the k to the po 阅读全文

摘要： Problem:http://uva.onlinejudge.org/external/5/541.htmlBasic Idea: calculate the sum of each row and each column respectively. If found any more than two row/column has an odd sum, or one row has odd sum, but all column have even sum, then it's corrupted; if one row & one column has old sum, 阅读全文

摘要： 2-1:2-2:a) loop invariant,initial condition, termination condition.b) A[j] is thesmallest element in A[j .. length[A]]. initialization: at the start of the loop, j = length[A].It's the smallest element in A[length[A] ~ length[A]] (because it'sthe only element). maintenance: if j = k is the s 阅读全文

摘要： 2.3-1:2.3-2:function MERGE-IMPROVED(A,p,q,r){ create array B[0 ... r - p]; int i = p, j = q+1, t =0; while(i<=q && j<=r){ if(A[i] >= A[j]){ B[t++] = A[j++]; } else{ B[t++] = A[i++]; } } if(i>q){ // i pile exhaustedwhile(j<=r) B[t++] = A... 阅读全文

摘要： 2.2-1:O(n^3)2.2-2:Selection-Sort(A): for( i=0 ; i<A.size()-1 ; i++){ max = A[i]; index = 0; for(j=i+1; j<A.size(); j++){ if(max < A[j]){ max = A[j]; index = j; } } // then swap the next max and the current element A[j] = A[i]; A[i] = max; }2.2-3:Say we're sorting nelementsWorst Case: nB 阅读全文

摘要： 2.1-1：2.1-2:NON-INCREASING-INSERTION-SORT(A) forj: 1~ length(A)-1: if(A[j]< A[j-1]): key= A[j]; t = j-1; while(A[t]< key && t>=0): A[t+1]= A[t]; t--; A[t]= key; returnA;2.1-3:// Assumption: Ifthere're duplicated elements, only return// the index of the first element.LINEAR-SEARC 阅读全文

摘要： 1.1-1Sorting: rank thestudent's according to the scoreDetermining thebest order for multiplying matrices: When doingprobability calculation, one may require lots of matrices multiplication for the cumulative probabilityresultsFinding the convexhull: 1.1-2Space1.1-3Linked list:Strength: can beuse 阅读全文

摘要： 虽然下周就考试了，还有很多没复习……我还如此淡定地花了整整一个下午+晚上想了一个实际问题……（其实也不是那么特别实际，但的确是来自于现实生活的DP问题）问题+讨论的URL：http://topic.csdn.net/u/20110421/17/dc8e63bf-7992-487f-a112-049cf390cb1c.html?seed=1122826158&r=72946293#r_72946293（因为考虑到很多人很懒，不想打开URL去看，我把里面重点的内容复制过来）问题陈述：比如CSDN的积分形式（以下为假设），每当你获得一次积分，服务器便保留这时的时间以及总分。现在要求在哪一段时 阅读全文

摘要： URL: http://uva.onlinejudge.org/external/107/10702.htmlAC Code:/*Problem ID: 10702Solver: songyyStart Time: End Time:Time Spent:*/#include <iostream>#include <stdio.h>#include <vector>#include <string.h>#define INF 1000000usingnamespace std;int graph[100][100];int memo[100][1 阅读全文