Hiking
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 431 Accepted Submission(s): 236
Special Judge
1. he selects a soda not invited before;
2. he tells soda the number of soda who agree to go hiking by now;
3. soda will agree or disagree according to the number he hears.
Note: beta will always tell the truth and soda will agree if and only if the number he hears is no less than li and no larger than ri, otherwise he will disagree. Once soda agrees to go hiking he will not regret even if the final total number fails to meet some soda's will.
Help beta design an invitation order that the number of soda who agree to go hiking is maximum.
The first contains an integer n (1≤n≤105), the number of soda. The second line constains n integers l1,l2,…,ln. The third line constains n integers r1,r2,…,rn. (0≤li≤ri≤n)
It is guaranteed that the total number of soda in the input doesn't exceed 1000000. The number of test cases in the input doesn't exceed 600.
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 100000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
struct node{
int l,r,id;
node(int l,int r,int id):l(l),r(r),id(id){};
};
struct cmp{//如过把运算符比较放在结构体里面 会更快些
bool operator()(node a,node b){
if(a.r==b.r)
return a.l>b.l;
return a.r>b.r;
}
};
int vis[maxn],a_id[maxn],l[maxn],r[maxn];
vector<node>v[maxn];
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
int t;
int n;
cin>>t;
while(t--){
cle(vis);
scanf("%d",&n);
for(int i=1;i<=n;i++)scanf("%d",&l[i]);
for(int i=0;i<=n;i++)v[i].clear();
for(int i=1;i<=n;i++)scanf("%d",&r[i]);
for(int i=1;i<=n;i++){
v[l[i]].push_back(node(l[i],r[i],i));
}
priority_queue<node,vector<node>,cmp>q;
int num=0;
for(int i=0;i<=n;i++){//枚举i
if(v[0].size()==0)break;//如果i不为0那么就不会又接下来的事情
for(int j=0;j<v[i].size();j++){
q.push(v[i][j]);
}//将左端点是i的都压入队列
if(q.empty())break;//队列为空 那么就没有满足条件的了
while(!q.empty()){
node x=q.top();
if(x.r>=i) {a_id[num++]=x.id;q.pop();vis[x.id]=1;break;}//找到最小的右端点
else q.pop();
}
}
if(num==0){
printf("%d\n",num);
for(int i=1;i<n;i++)
printf("%d ",i);
printf("%d\n",n);
}
else{
printf("%d\n",num);
printf("%d",a_id[0]);
for(int i=1;i<num;i++)printf(" %d",a_id[i]);
for(int i=1;i<=n;i++)if(!vis[i])printf(" %d",i);
printf("\n");
}
}
return 0;
}