Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 177    Accepted Submission(s): 70


Problem Description
科学家小沃沃在研究病毒传播的规律,从而控制疫情。

有 t 以后,这个人也会沾上荧光粉。

从小到大输出实验结束后身体上有荧光粉的人的编号。
 

 

Input
第一行一个整数 200000
 

 

Output
对于每组数据输出一行,表示所有患者的编号。按编号从小到大输出。
 

 

Sample Input
2 4 2 1 1 2 2 3 2 2 3 3 4 4 1 4 4 1 2 1 3 3 1 1 3 1 6 1 3 4 1 5 1 6 1 1 5 1
 

 

Sample Output
1 2 3 1 2 样例解释 Case 1: 第 2 时刻,位置 2,1 与 2 相遇,2 沾上了。 第 4 时刻,位置 4,2 与 3 相遇,3 沾上了。
 

 

Source
 

 

Recommend
heyang
 

 

Statistic | Submit | Discuss | Note

析:直接按照时间t进行模拟就行了,把有过感染的标记一下就OK了。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#include <unordered_map>
#define debug() puts("++++")
#define print(x) cout<<(x)<<endl
// #define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e4 + 7;
const int maxm = 2000000 + 7;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
int n, m;

inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){
  int x;  cin >> x;  return x;
}


bool a[maxn];
map<P, std::vector<int> > mp;

int main(){
  int T;  scanf("%d", &T);
  while(T--){
    scanf("%d", &n);  ms(a, 0);  mp.cl;
    a[1] = true;
    for(int i = 1; i <= n; ++i){
      scanf("%d", &m);
      for(int j = 0; j < m; ++j){
        int t, p;  scanf("%d %d", &t, &p);
        mp[P(t, p)].pb(i);
      }
    }
    for(auto &it: mp){
      std::vector<int> &v = it.se;
      bool ok = false;
      for(auto &x : v){
        if(a[x]){
          ok = true;
          break;
        }
      }
      if(ok){
        for(auto &x : v)  a[x] = true;
      }
    }
    printf("1");
    for(int i = 2; i <= n; ++i)  
      if(a[i]) printf(" %d", i);
    printf("\n");
    
  }
  return 0;
}

  

相关文章: