职务地址:HDU 3729 二分图最大匹配+按字典序输出结果. 仅仅要从数字大的開始匹配就能够保证字典序最大了.群里有人问. . 就顺手写了这题. . 代码例如以下: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int vis[110000], head[110000], cnt, link[110000]…

I - I'm Telling the Truth Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVALive 5033 Description After this year's college-entrance exam, the teacher did a survey in his class on students' score. There…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3729 I'm Telling the Truth Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1700    Accepted Submission(s): 853 Problem Description After this year’…

http://acm.hdu.edu.cn/showproblem.php?pid=3729 I'm Telling the Truth Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1427    Accepted Submission(s): 719 Problem Description After this year’s col…

.I'm Telling the Truth Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1740 Accepted Submission(s): 871 Problem Description After this year's college-entrance exam, the teacher did a survey in his…

I'm Telling the Truth Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2006    Accepted Submission(s): 1011 Problem Description After this year’s college-entrance exam, the teacher did a survey i…

裸的二分图匹配.需要输出方案. #include<cstdio> #include<cstring> #include<vector> #include<algorithm> #include<iostream> using namespace std; #define M 100005 #define N 65 bool vis[M]; vector<int> g[N]; int now[M]; int n,m; int dfs(i…

题意:给定 n 个人成绩排名区间,然后问你最多有多少人成绩是真实的. 析:真是没想到二分匹配,....后来看到,一下子就明白了,原来是水题,二分匹配,只要把每个人和他对应的区间连起来就好,跑一次二分匹配,裸的. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #inc…

思路 题意:该题主要说几个同学分别说出自己的名次所处区间,最后输出可能存在的未说谎的人数及对应的学生编号,而且要求字典序最大. 思路:刚刚接触匈牙利算法,了解的还不太清楚,附一个专门讲解匈牙利算法的博文,个人认为讲的比较清晰. AC代码 #include<iostream> #include<cstdio> #include<cstring> using namespace std; int T, n; struct Stue { int l, r; }; Stue p…

题意:有n个人,每个人给出自己的名次区间,问最多有多少个人没撒谎,如果有多解,输出字典序最大的解. 分析: 1.因为字典序最大,所以从后往前分析. 2.假设后面的人没说谎,并将此作为已知条件,然后从后往前依次给每个人找到合适的名次,输出所有能找到合适名次的人即可. 3.假定给第i个人安排名次,第i+1~n个人名次已经安排好,假如第i个人想占的名次被第j个人所占,那就从第j个人可以占的名次中再找个合适的名次给j,然后把空出来的这个名次给i,如果i可以占的所有名次都被占且占领的人找不到其他可以占的名…