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’…

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…

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: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…

http://acm.hdu.edu.cn/showproblem.php?pid=2236 题意:中文题意. 思路:先找出最大和最小值,然后二分差值,对于每一个差值从下界开始枚举判断能不能二分匹配. #include <cstdio> #include <algorithm> #include <iostream> #include <cstring> #include <string> #include <cmath> #incl…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1498 50 years, 50 colors Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1918    Accepted Submission(s): 1058 Problem Description On Octorber 21st,…

Rain on your Parade Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 655350/165535 K (Java/Others)Total Submission(s): 2154    Accepted Submission(s): 662 Problem Description You’re giving a party in the garden of your villa by the sea. The pa…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2063 二分匹配最大匹配数简单题,匈牙利算法.学习二分匹配传送门:http://blog.csdn.net/dark_scope/article/details/8880547 #include <iostream> #include <cstdio> #include <cstring> #include <vector> using namespace std…

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]…