看到大神说location的值不会超过1000,所以这就简单很多了,用一个deg数组记录下来每个点的度,出度-1,入读+1这样. 最后判断每个点的度是否为0即可. 至于为什么会这样,据说是套数据套出来的,比如在代码里加一句if(a >= 1000) for(;;),get新技能! 如果按正常来做的话,我能想到的就是遍历map了. #include <cstdio> #include <cstring> ; int deg[maxn]; int main() { //freop…

题意:给出很多对数字,看看每一对(a,b)能不能找到对应的(b,a). 放在贪心这其实有点像检索. 用stl做,map+pair. 记录每一对出现的次数,然后遍历看看对应的那一对出现的次数有没有和自己出现的此时一样即可. 代码: /* * Author: illuz <iilluzen@gmail.com> * Blog: http://blog.csdn.net/hcbbt * File: uva10763.cpp * Lauguage: C/C++ * Create Date: 2013-…

题目连接:10763 Foreign Exchange 题目大意:给出交换学生的原先国家和所去的国家,交换成功的条件是如果A国给B国一个学生,对应的B国也必须给A国一个学生,否则就是交换失败. 解题思路: 给出数据  10 x  y 1 2 2 1 3 4 4 3 100 200 200 100 57 2 2 57 1 2 2 1 按照排序: xy y x 12 1 2 12 1 2 21 2 1 21 2 1 257 2 57   34 3 4  43 4 3 572 57 2 100 200…

题目:10763 - Foreign Exchange 题目大意:给出每一个同学想要的交换坐标 a, b 代表这位同学在位置a希望能和b位置的同学交换.要求每一位同学都能找到和他交换的交换生. 解题思路:把给定的原先给定的序列,交换前后位置后得到新的序列. 假设这个新的序列和原来的序列同样就符合要求.由于  (a.b) (b. a)若是成对出现.那么前后交换后还是(b. a)(a.b). 代码: #include <stdio.h> #include <string.h> #inc…

Foreign Exchange Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over the last few years, demand has sky-rocketed and now you need assistance w…

Foreign Exchange Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over th…

 10763 Foreign ExchangeYour non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinatesa very successful foreign student exchange program. Over the last few years, demand hassky-rocketed and now you need assistan…

 Foreign Exchange Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over the last few years, demand has sky-rocketed and now you need assistance…

题意:有若干交换生.若干学校,有人希望从A校到B校,有的想从B到C.C到A等等等等.如果有人想从A到B也刚好有人想从B到A,那么可以交换(不允许一对多.多对一).看作后如果有人找不到人交换,那么整个交换计划失败. 代码:(Accepted, 50ms) //UVa10763 - Foreign Exchange #include<cstdio> #include<cstring> int N, a, b,all[1003][1003]; int main() { //freopen…

  Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu   Description Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over the…

  Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over the last few years, demand has sky-rocketed and now you need assistance with your task.…

Your non-profitorganization (iCORE - international Confederationof Revolver Enthusiasts) coordinates a very successfulforeign student exchange program. Over the last few years, demand hassky-rocketed and now you need assistance with your task. The pr…

题目 题目     分析 没什么好说的,字符串拼接一下再放进map.其实可以直接开俩数组排序后对比一下,但是我还是想熟悉熟悉map用法. 呃400ms,有点慢.     代码 #include <bits/stdc++.h> using namespace std; int read(string a) { int ans=0,i=0; while(i<a.length()){ans=(ans<<3)+(ans<<1)+a[i]-'0';i++;} return…

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1332 #include<iostream> #include<stdio.h> #include<string.h> #include<string> #include<map> using namespace std; stri…

题目链接: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3148 题意:给K个数组,每个数组含有K个整数,从每个数组中各选一个数加起来,得到一个sum,这样的选法一共有K^k种,现在求这样的和中最小的K个sum. 解法:先考虑两个数组的情形 假设A1<=A2<=`````<=AK B1<=B2<=`````&…

10763:水题不解释直接贴代码. #include <iostream> #include <string.h> #include <stdio.h> #include <algorithm> #include <math.h> #include <queue> #define eps 1e-9 typedef long long ll; using namespace std; int n; ]; int main() { int…

There is a bag-like data structure, supporting two operations: 1 x Throw an element x into the bag. 2 Take out an element from the bag. Given a sequence of operations with return values, you're going to guess the data structure. It is a stack (Last-I…

Queues and Priority Queues are data structures which are known to most computer scientists. The Team Queue, however, is not so well known, though it occurs often in everyday life. At lunch time the queue in front of the Mensa is a team queue, for exa…

题意: 有一个集合栈计算机,栈中的元素全部是集合,还有一些相关的操作.输出每次操作后栈顶集合元素的个数. 分析: 这个题感觉有点抽象,集合还能套集合,倒是和题中配的套娃那个图很贴切. 把集合映射成ID,就可以用 stack<int>来模拟题中的集合栈,然后用 vector<Set> 来根据下标进行集合的索引. 代码虽短,但还须多体会. #include <cstdio> #include <string> #include <vector> #i…

题意: 作图为n个建筑物的俯视图,右图为从南向北看的正视图,按从左往右的顺序输出可见建筑物的标号. 分析: 题中已经说了,要么x相同,要么x相差足够大,不会出现精度问题. 给这n个建筑物从左往右排序,每个建筑物的两个端点,排序去重以后可以得到m个相邻的小区间.枚举这些区间,判断建筑物是否可见. 离散化刚开始接触这个词,感觉十分高冷.现在来看倒是很形象,因为是浮点数,所以不可能枚举所有的横坐标,但可以分割成若干的小区间,这个进行判断.即:将无限变为有限. 再说一下unique函数,调用之前必须先排…

题目描述: 解题思路: 开一个数组,读入一次交换两个数,如果最后数组不变,即符合匹配 #include<iostream> #include<cstdio> #include<string> #include<algorithm> #define MAXN 500005 using namespace std; int arr[MAXN]; void swap(int a,int b){ int t=arr[a]; arr[a] = arr[b]; arr[…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 如果x>y 则num[(x,y)]--; 否则num[(x,y)]++; 看看每一个二元组的num值是不是都为0就好. [代码] #include <bits/stdc++.h> using namespace std; int n; map <pair<int, int>, int> mmap; int main() { //freopen("F:\\rush.txt", &…

题目链接: https://www.luogu.org/problemnew/show/UVA10763 题目分析: 本题我首先想到的做法是把每一个数都map一下,然后互相判断,例如a,b两人准备交换,那么m[a]=b,m[b]=a,最后再判断如果m[m[a]]=a就行,但是因为一个学生在双方都同意的情况下可以先后与多个学生交换,所以我的做法是: 先把每对学生都按照先小后大的顺序排好 if(a[i].x>a[i].y)swap(a[i].x,a[i].y); 然后把所以学生交换对都进行排序 so…

例题5--9 数据库 Database UVa 1592 #include<iostream> #include<stdio.h> #include<string.h> #include<cmath> #include<string> #include<queue> #include<stack> #include<vector> #include<map> #include<set>…

题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics 10300 - Ecological Premium 458 - The Decoder 494 - Kindergarten Counting Game 414 - Machined Surfaces 490 - Rotating Sentences 445 - Marvelous Mazes…

MENU: Transactions > Financial > Revalue Open Currency Balances 使用频率: - 每个月月底,结账前, 手工操作. - 或者在Period Close Checklist会有一步需要操作:Revalue Open Foreign Currency Balances Note: - 页面Month End Currency Revaluation,每个Subsidiary都要做,在列表下面系统会列出所有等待有可能存在交易外币金额的科目…

PDMS RvmTranslator eryar@163.com Abstract. AVEVA Review is used for 3D model visualisation for plant or ship design, construction and operation. The RVM file is main input of AVEVA Review. RvmTranslator can translate RVM file to STEP, IGES, STL, etc…

Finance knowledge Trading---At the core of our business model is Trading, which involves the buying and selling of financial tools to generate profit. Trading takes place in our Global Markets division, which spans collateralised financing, commoditi…

英语新闻常用词汇与短语 经济篇 accumulated deficit 累计赤字 active trade balance 贸易顺差 adverse trade balance 贸易逆差 aid 援助 allocation of funds 资金分配 allotment 拨款 allowance/grant/subsidy 补贴,补助金,津贴 amortization 摊销,摊还,分期偿付 annuity 年金 article 物品,商品 assigned 过户 autarchy 闭关自守 av…

* 设置->技术->数据结构->模型                模型    模型描述    类型    瞬态模型account.account    科目    基础对象    假account.account.template    科目模板    基础对象    假account.account.type    类型    基础对象    假account.addtmpl.wizard    account.addtmpl.wizard    基础对象    真account.a…