hdu 3729(二分图最大匹配)
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
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.
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.
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)
4
5004 5005
5005 5006
5004 5006
5004 5006
7
4 5
2 3
1 2
2 2
4 4
2 3
3 4
2 3 4
5
1 3 5 6 7
#include<iostream>
#include<cstdio>
#include<cstring>
#include <algorithm>
#include <math.h>
#include <queue>
using namespace std;
const int INF = ;
const int N = ;
const int M = ;
int graph[N][M];
struct Rank{
int l,r;
}r[N];
int ans[N];
int n;
int linker[M];
bool vis[M];
bool dfs(int u){
for(int v = r[u].l;v<=r[u].r;v++){
if(!vis[v]&&graph[u][v]){
vis[v] = true;
if(linker[v]==-||dfs(linker[v])){
linker[v] = u;
return true;
}
}
}
return false;
}
int main()
{
int tcase;
scanf("%d",&tcase);
while(tcase--){
memset(graph,,sizeof(graph));
scanf("%d",&n);
for(int i=;i<=n;i++){
int a,b;
scanf("%d%d",&a,&b);
r[i].l = a,r[i].r = b;
for(int j=a;j<=b;j++){
graph[i][j] = ;
}
}
memset(linker,-,sizeof(linker));
int id = ,res=;
for(int i=n;i>=;i--){
memset(vis,false,sizeof(vis));
if(dfs(i)){
ans[id++] = i;
res++;
}
}
printf("%d\n",res);
for(int i=id-;i>=;i--){
printf("%d ",ans[i]);
}
printf("%d\n",ans[]);
}
return ;
}
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