1.repo start <topic_name> 开启一个新的主题,其实就是每个Project都新建一个分支. repo start newbranchname . 创建新的branch分支. "." 代表当前工作的branch 分支. 2.repo init -u <url> [OPTIONS] repo init -u git://android.git.kernel.org/platform/manifest.git repo init -u URL 用

题目传送门 题意:求单词的最长公共子序列,并要求打印路径 分析:LCS 将单词看成一个点,dp[i][j] = dp[i-1][j-1] + 1 (s1[i] == s2[j]), dp[i][j] = max (dp[i-1][j], dp[i][j-1]) 代码: #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <strin

#include<stdio.h> #include<string.h> #include<stdlib.h> #include<ctype.h> #include<algorithm> #define N 1010 using namespace std; int dp[N], path[N][N], w[N]; int main() { int v, n; while(~scanf("%d", &v)) { sca

题目链接: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=995 Problem D: The Necklace  My little sister had a beautiful necklace made of colorful beads. Two successive beads in the necklace shared a c

Scandinavians often make vacation during the Easter holidays in the largest ski resort Are. Are provides fantastic ski conditions, many ski lifts and slopes of various difficulty profiles. However, some lifts go faster than others, and some are so po

题目描述 一个无环的有向图称为无环图(Directed Acyclic Graph),简称DAG图.     AOE(Activity On Edge)网:顾名思义,用边表示活动的网,当然它也是DAG.与AOV不同,活动都表示在了边上,如下图所示:                                         如上所示,共有11项活动(11条边),9个事件(9个顶点).整个工程只有一个开始点和一个完成点.即只有一个入度为零的点(源点)和只有一个出度为零的点(汇点).    关键

题意: 给你两个空瓶子,只有三种操作 一.把一个瓶子灌满 二.把一个瓶子清空 三.把一个瓶子里面的水灌到另一个瓶子里面去(倒满之后要是还存在水那就依然在那个瓶子里面,或者被灌的瓶子有可能没满) 思路:BFS,打印路径时需技巧. //G++ 840K 0MS #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <stack> #in

推荐一篇炒鸡赞的blog. 以下代码中有打印路径. #include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <queue> #include <cmath> #include <stack> #include <map> #include <

  Compromise  In a few months the European Currency Union will become a reality. However, to join the club, the Maastricht criteria must be fulfilled, and this is not a trivial task for the countries (maybe except for Luxembourg). To enforce that Ger

链接:UVa 103 题意:给n维图形,它们的边长是{d1,d2,d3...dn},  对于两个n维图形,求满足当中一个的全部边长 依照随意顺序都一一相应小于还有一个的边长,这种最长序列的个数,而且打印随意一个最长子串的路径, 比如:a(9,5,7,3),b(6,10,8,2),c(9,7,5,1),a和b不满足,但c和b满足 分析:首先对没组边长从小到大排序,再对各组图形按最小边排序,再求最大子串, 对于打印路径,能够逆序循环,也可递归求解 #include<cstdio> #include

option=com_onlinejudge&Itemid=8&page=show_problem&problem=1072">链接:UVa 10131 题意:给定若干大象的体重及智商值.求满足大象体重严格递增,智商严格递减的序列的最大个数. 并打印随意一组取得最大值的序列的大象编号 分析:这个是LIS的应用,仅仅只是推断条件有两个,能够先对大象的体重排序,可是要打印路径. 那就必须得回溯求路径.能够直接逆序循环求,当然递归也是一个好的选择 #include<

早晨碰到了一题挺裸的最短路问题需要打印路径:vijos1635 1.首先说说spfa的方法: 其实自己之前打的最多的spfa是在网格上的那种,也就是二维的 一维的需要邻接表+queue 以及对于queue的操作,自己也是醉了 这里贴一个模板(不含打印路径): #include<cstdio> #include<cstring> #include<queue> #include<iostream> using namespace std; const int

Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18847    Accepted Submission(s): 6090Special Judge Problem Description The Princess has been abducted by the BEelzebub

给出两个字符串A B,求A与B的最长公共子序列(子序列不要求是连续的).   比如两个串为:   abcicba abdkscab   ab是两个串的子序列,abc也是,abca也是,其中abca是这两个字符串最长的子序列. Input 第1行:字符串A 第2行:字符串B (A,B的长度 <= 1000) Output 输出最长的子序列,如果有多个,随意输出1个. Input示例 abcicba abdkscab Output示例 abca #include <bits/stdc++.h>

FatMouse believes that the fatter a mouse is, the faster it runs. To disprove this, you want to take the data on a collection of mice and put as large a subset of this data as possible into a sequence so that the weights are increasing, but the speed

Problem Description FatMouse believes that the fatter a mouse is, the faster it runs. To disprove this, you want to take the data on a collection of mice and put as large a subset of this data as possible into a sequence so that the weights are incre

本文转载自:https://blog.csdn.net/u010164190/article/details/78332484 .repo forall命令 # repo forall -help # repo forall -c: 此命令遍历所有的git仓库,并在每个仓库执行-c所指定的命令,被执行的命令不限于git命令,而是任何被系统支持的命令,比如:ls, git log, git status等 .repo forall -c使用 # 切换分支 # repo formal -c git

题目链接(貌似未报名的不能进去):https://www.nowcoder.com/acm/contest/141/A 题目: 题意:背包题意,并打印路径. 思路:正常背包思路,不过五维的dp很容易爆内存,比赛时无限爆,后面队友提醒用short就过了.不过也可以用滚动减少内存消耗,两种代码实现都贴一下吧~ 五维代码实现如下: #include <set> #include <map> #include <queue> #include <stack> #in

uva 紫书例题,这个区间dp最容易错的应该是(S)这种匹配情况,如果不是题目中给了提示我就忽略了,只想着左右分割忘记了这种特殊的例子. dp[i][j]=MIN{dp[i+1][j-1] | if(match(i,j) , dp[i][k]+dp[k+1][j] | i<=k<=j .}注意初始化dp[i][i]=1,表示1个字符最少需要一个才能匹配,dp[i+1][i]=0,因为可能只有两个字符使得i+1>j-1,我们可以认为中间是空字符已经匹配了. 打印路径利用了递归,很巧妙,lr

/* LCS */ #include<bits/stdc++.h> using namespace std; const int maxn = 1000; int dp[maxn][maxn], c[maxn][maxn]; int str1[maxn],str2[maxn]; int k; void dfs(int i,int j){ //打印路径 if(i == 0 || j == 0) return; if(c[i][j] == 1){ dfs(i-1,j-1); k--; printf

A. Shortest path of the king time limit per test 1 second memory limit per test 64 megabytes input standard input output standard output The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has busi