I'm Telling the Truth（二分匹配）

               

I'm Telling the Truth

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1370    Accepted Submission(s): 696

Problem Description

After this year’s college-entrance exam, the teacher did a survey in his class on students’ score. There are n students in the class. The students didn’t want to tell their teacher their exact score; they only told their teacher their rank in the province (in the form of intervals).

After asking all the students, the teacher found that some students didn’t tell the truth. For example, Student1 said he was between 5004th and 5005th, Student2 said he was between 5005th and 5006th, Student3 said he was between 5004th and 5006th, Student4 said he was between 5004th and 5006th, too. This situation is obviously impossible. So at least one told a lie. Because the teacher thinks most of his students are honest, he wants to know how many students told the truth at most.

 

Input

There is an integer in the first line, represents the number of cases (at most 100 cases). In the first line of every case, an integer n (n <= 60) represents the number of students. In the next n lines of every case, there are 2 numbers in each line, X_i and Y_i (1 <= X_i <= Y_i <= 100000), means the i-th student’s rank is between X_i and Y_i, inclusive.

 

Output

Output 2 lines for every case. Output a single number in the first line, which means the number of students who told the truth at most. In the second line, output the students who tell the truth, separated by a space. Please note that there are no spaces at the head or tail of each line. If there are more than one way, output the list with maximum lexicographic. (In the example above, 1 2 3;1 2 4;1 3 4;2 3 4 are all OK, and 2 3 4 with maximum lexicographic)

 

Sample Input

245004 50055005 50065004 50065004 500674 52 31 22 24 42 33 4

 

Sample Output

32 3 451 3 5 6 7

PS：自己的第一道二分图匹配，觉得这题作为二分匹配的入门指导题不错，

写下题解，增加自己对二分图的最大匹配的理解吧！

题意： 题目是老师想知道他的学生在这次的考试考得咋样，于是他就向他的学生提问，然而他的学生都不想告诉他他们的成绩，只肯说他们在级里的排名，并且有些学生会说谎，谎报排名等级，于是老师想知道究竟有多少个人没有说谎，因为老师很相信他的学生，所以他想知道最多的人说实话的人数；

意解： 题目是想找出最大的匹配数，所以可以用二分图的最大匹配的思想来解题，具体可以用匈牙利算法解。每次可以在一个区间中找出可以匹配的等级，找不到就递归继续查找；最大匹配： 首先可以把学生的编号作为一个点集，把学生的排名等级作为另一个等级，然后就可以构造最大匹配了。

PS：一个二分图中的最大匹配数等于这个图中的最小点覆盖数

König定理是一个二分图中很重要的定理，它的意思是，一个二分图中的最大匹配数等于这个图中的最小点覆盖数。如果你还不知道什么是最小点覆盖，我也在这里说一下：假如选了一个点就相当于覆盖了以它为端点的所有边，你需要选择最少的点来覆盖所有的边。

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <set>#include <queue>#include <vector>#include <cstdlib>#include <algorithm>#define ls u << 1#define rs u << 1 | 1#define lson l, mid, u << 1#define rson mid + 1, r, u << 1 | 1using namespace std;const int M = 1e5 + 5;int x[66],y[66],vis[M],line[M];/***************************//* 二分图的最大匹配/* 人的编号和排名构成二分图/* 题目因为要按最大的字典序输出/* 所以可以从后面往前匹配/* 递归搜索可以匹配的边/***************************/bool find(int n){    for(int i = x[n]; i <= y[n]; i++)    {        if(!vis[i])        {            vis[i] = 1;            if(!line[i] || find(line[i]))  //如果当前的等级已经被匹配，则递归搜索还未匹配的等级            {                line[i] = n; //把当前的人和等级连一条边；                return true;            }        }    }    return false;}int main(){//    freopen("in.txt","r",stdin);//    freopen("out.txt","w",stdout);    int t,n;    scanf("%d",&t);    while(t--)    {        fill(line,line + M, 0);        int ans[66],res = 0;        scanf("%d",&n);        for(int i = 0; i < n; i++)        {            scanf("%d %d",x + i, y + i);        }        for(int i = n - 1; i >= 0; i--)        {            fill(vis,vis + M, 0); //记得初始化辅助数组，记录当前的等级是否已经被匹配；                        if(find(i)) ans[res++] = i + 1; //找到匹配的入数组；        }        printf("%d\n", res);        for(int i = res - 1; i >= 0; i--)        {            if(i == 0) printf("%d\n",ans[i]);            else printf("%d ",ans[i]);        }    }        return 0;}