fyfcauc的专栏

#include #include #include struct operatorParam{    char operand;    char operandRes;    struct operatorParam * left;    //For p q r s t and N, only one param.    struct operatorParam

#include #include struct gridCell{    char direction;    bool passed;    int step;};typedef struct gridCell gridCell;typedef struct moveRecord moveRecord;void moveRobot(gridC

/* 题意： 对于每一个月来说，如果盈利则盈利S，如果亏空则亏d。每五个月进行一次统计，共统计八次（1-5月一次，2-6月一次.......） 统计的结果是这八次都是亏空。问题：判断全年是否能盈利，如果能则求出最大的盈利。如果不能盈利则输出Deficit MS公司每月要么盈余，要么亏损，且一年中每个月的盈余是一样的，亏损也是一样的。财务统计是每五个月统计一次收入总额，即1-

重新复习了算法导论的DP一章：以前也看过， 不过那时候就是看， 基本没怎么用，现在回过头来再看，还是很多发现。根据求解子问题顺序的不同， 程序也会变为递归和迭代两种， 比如 a[i, k](k == 1 || k ==2, k依赖于每一步的选择空间有几个) 的结果依赖于 a[i-1, 1] 和 a[i-1, 2]。如果是顺序的从 i = 0 开始求， 那么就是迭代。如果是逆序的

#include #include int getMaxSum(int *array, int beginPos, int endPos, int *dpArray) {    int MaxBegin = beginPos;    int MaxEnd = beginPos;    int curMax = array[beginPos];    int curMax

#include #include #include #include struct pointInfo {    int x;    int y;    int havl;    struct pointInfo * next;};struct point {    int x;    int y;};typedef struc

//1380K 47MS G++#include #include #define MAX 510int N;int K;int Vset1[MAX]; // rowint Vset2[MAX]; // columnint G[MAX][MAX]; // 0: no asteriod, 1: asteriodint assignedFlag[MAX];int waitin

#include #include #define MAX 35#define GRID_MAX MAX*MAXint G[MAX][MAX];char G_relation[GRID_MAX][GRID_MAX];int M; // lengthint N; // heightint K;int validGridNum;int V1[GRID

#include #include #include struct pointInfo {    int x;    int y;    int havl;    char used;    char usedAsSquare;};struct point {    int x;    int y;};typedef struct

From： http://blog.youkuaiyun.com/wall_f/article/details/8187144鉴于我一直没有分清楚最小边覆盖与最小路径覆盖的关系，于是我就写写总结。边覆盖集：通俗地讲，所谓边覆盖集，就是G中所有的顶点都是E*中某条边的邻接顶点（边覆盖顶点），一条边只能覆盖2个顶点。注意：在无向图中存在用尽量少的边去“覆盖”住所有的顶点，所以边覆盖集有极小与最小的

//440K 0MS G++#include #include #define MAX 125int G[MAX][MAX];int iNum;int sNum;int V1[MAX];int V2[MAX];char waitingMove[MAX];char getPair(int checkId) { for (int i = 1; i <= iNum

#include #include #include using namespace std;priority_queue, greater > intQueue;void getMinCost() {    long long totalCost = 0;    while(intQueue.size() > 1) {        long long

// 936K 32MS G++#include #include #include #define MAX 210#define INF 999999long long intersectionMap[MAX][MAX];int prev[MAX];int flow[MAX];int N;int M; // intersection numusing

#include #include #include #include #include #include using namespace std;struct dict_entry{    int hval;    char foreign[15];    char local[15];    struct dict_entry * next;

From: https://www.byvoid.com/blog/match-km/[二分图带权匹配与最佳匹配]什么是二分图的带权匹配？二分图的带权匹配就是求出一个匹配集合，使得集合中边的权值之和最大或最小。而二分图的最佳匹配则一定为完备匹配，在此基础上，才要求匹配的边权值之和最大或最小。二分图的带权匹配与最佳匹配不等价，也不互相包含。我们可以使用KM算法实现求二分图的最

#include #include #include #include #include #include using namespace std;const int CHAR_NUM = 26;struct tireNode{    char val[22];    struct tireNode * tireChildNode[CHAR_NU

#include // #include #include // #include // #include #include #define CHAR_NUM 10#define STATIC_CAP 700000struct tireNode{    int childPos[CHAR_NUM];    int appearTime;   

// 2712K 266MS G++ #include #include #include using namespace std;#define MAX 520int G[MAX][MAX];int G_F[MAX][MAX];int iNum;int sNum;int V1[MAX];int V2[MAX];char waitingMove[MAX];

#include #include #include #define HEAP_SIZE 30010template class Less {public:    char operator()(const T & p1, const T & p2) const {        return p1     }};template cla

#include #include #include #define MAX_SUBSTRING_NUM 1600000#define BIG_PRIME 1000007#define MAX_N 25char baseString[3000000];struct NodeInfo{    char val[MAX_N];    int next;}

#include #include struct fishingOP {    int maxFishing;    int stayInterval;};typedef struct fishingOP fishingOP;void getMostFish(int * fi, int * di, int * ti, int lakeNum, int i

#include #include #include using namespace std;#define COW_NUM 50000class cow_vote {public:    int index;    long Avote;    long Bvote;    void operator=(const cow_vote & p)

#include #include #include #include using namespace std;int tmpArray[300];int sticks[64];char stickUsed[64];int maxLevel;int comp( const void *a, const void * b){return *

#include #include int alignmentMatix[100][100] = {0};int DPResult[101][101];void init() {    alignmentMatix['A']['A'] = 5;    alignmentMatix['A']['C'] = -1;    alignmentMatix['A']['G

#include #include #include using namespace std;#define MAX 110int map[MAX][MAX]; // 0: space, even: human odd: houseint N; // number of rows, heightint M; // number of columns. widthstruc

#include int getArrayMaxSum(int *Array, int length) {    // printf("======================================\n");    // for (int i = 0; i     //     printf("%d ", Array[i]);    // }  

#include #include struct pearlInfo{    int number;    int price;};typedef struct pearlInfo pearlInfo;struct BuyOP {    int cost;    int higerGlass;};typedef struct BuyO

// 700K 0MS G++#include #include #include using namespace std;#define MAX 50char maze[MAX][MAX];char leftFirst[2] = {-1, 1};char rightFirst[2] = {1, -1};struct Point { int x; int y; i

// 1152K 47MS G++#include #include #include #define MAX 100001using namespace std;struct Pos { int x; int time;};typedef struct Pos Pos;queue BFSQueue;int Bx;int Ex;in

#include int UnimodalNum[1000][1000] = {0};void init() {    for(int i = 0; i         for (int j = 999; j >= 0; j--) {            if (i == 0) { // N is 0.                 UnimodalNum[

//376K 172MS G++#include #include #define MAX 30int W;int H;int board[MAX][MAX];int Sx;int Sy;int Gx;int Gy;#define INF 99999int dfs(int curX, int curY, int throwTime) { int minThr

#include #include struct pearlInfo{    int number;    int price;};typedef struct pearlInfo pearlInfo; struct BuyOP {    int cost;    int higerGlass;};typedef struct Buy

//788K 0MS G++#include #include #include #define MAX 50using namespace std;char maze[MAX][MAX][MAX]; // z , y, xstruct Pos { int x; int y; int z; int time;};typedef struc

// 2604K 32MS G++ #include #include #include #include #include using namespace std;#define MAX 230map deviceList;int receptacleNum;int deviceNum;int adapterNum;struct flowNode { int

//376K 16MS G++#include #include #define MAX 28 int board[MAX][MAX];int p;int q;struct Point{ int x; int y;};typedef struct Point Point;Point visitOrder[MAX];int visitedSquareNum;in

// 376K 16MS G++#include #include #define MAX 10char occupiedColumn[MAX];char occupiedRow[MAX];char board[MAX][MAX];int W;int H;int chessNum;#define INF 99999int dfs(int rowId, int

#include #include double eForExp = 2.718281828;double getT(double D, double H) {    double h = 0.5555 * (6.11 * pow(eForExp, (5417.7530 * ((1/273.16) - (1/(D + 273.16))))) - 10.0);    pr

//372K 0MS G++#include #include #include #define BEGIN 'A'int graph[26][26];int repeaterUsingChannel[26];int channelUsedInfo[27]; // channel 1 ~ 27int totalChanne

// 1st AC 30052K 4313MS G++#include // #include #include #include #include using namespace std;const int CHAR_NUM = 26;struct tireNode {    // char val[20];    int tireChil

from http://blog.youkuaiyun.com/dm_vincent/article/details/7655764本文主要介绍解决动态连通性一类问题的一种算法，使用到了一种叫做并查集的数据结构，称为Union-Find。更多的信息可以参考Algorithms 一书的Section 1.5，实际上本文也就是基于它的一篇读后感吧。原文中更多的是给出一些结论，我尝试给出