HDU - 3729 I‘m Telling the Truth(二分图最大匹配)

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I'm Telling the Truth

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2448    Accepted Submission(s): 1204

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, Xi and Yi (1 <= Xi <= Yi <= 100000), means the i-th student’s rank is between Xi and Yi, 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

2 4 5004 5005 5005 5006 5004 5006 5004 5006 7 4 5 2 3 1 2 2 2 4 4 2 3 3 4

Sample Output

3 2 3 4 5 1 3 5 6 7

Source

2010 Asia Tianjin Regional Contest

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题目大意：

n个人给出自己的名次区间，可真可假，求出最多有多少人说实话。输出字典序最大。

思路：

做这个题之前匈牙利算法接触的很少，所以拿到题后就开始一直贪心（TAT）,结果就一直WA。

把每个人看作X集，区间看作Y集，将点与区间内的每一个数连边，求最大匹配数。从N->1扫描使得结果字典序最大。

附上AC代码：

#include<iostream>
#include<cstring>

using namespace std;
const int maxn=100000+5;
int vis[maxn],link[maxn],mark[maxn];
int T,n,sum,flag;

struct edges{
    int x,y;
}edge[maxn];

int dfs(int t){
    for(int i=edge[t].x;i<=edge[t].y;i++){
        if(mark[i]==0){
            mark[i]=1;
            if(link[i]==0||dfs(link[i])){
                link[i]=t;
                vis[t]=1;
                return 1;
            }
        }
    }
    return 0;
}

int main(){
    ios::sync_with_stdio(false);
    cin>>T;
    while(T--){
        sum=0,flag=0;
        memset(vis,0,sizeof(vis));
        memset(link,0,sizeof(link));
        cin>>n;
        for(int i=1;i<=n;i++)
            cin>>edge[i].x>>edge[i].y;
        for(int i=n;i>=1;i--){
            memset(mark,0,sizeof(mark));
            if(dfs(i)) sum++;
        }
        cout<<sum<<endl;
        for(int i=1;i<=n;i++){
            if(vis[i]&&flag==0){
                flag=1;
                cout<<i;
            }
            else if(vis[i]){
                cout<<' '<<i;
            }
        }
        cout<<endl;
    }
    return 0;
}