标签:
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1400 Accepted Submission(s): 875
1 #include <queue> 2 #include <cstdio> 3 #include <iostream> 4 using namespace std; 5 6 struct play 7 { 8 int posi, dy; 9 friend bool operator < (play posi, play dy) 10 { 11 if(posi.posi == dy.posi) 12 return posi.dy > dy.dy; //小 → → 大; 13 else 14 return posi.posi > dy.posi; //小 → → 大; 15 } 16 } ; 17 18 int main() 19 { 20 int t; 21 scanf("%d", &t); 22 while(t--) 23 { 24 priority_queue <play> q; 25 int i, m; 26 play t; 27 scanf("%d", &m); 28 for(i=0; i<m; i++) 29 { 30 scanf("%d %d", &t.posi, &t.dy); 31 q.push(t); 32 } 33 int ans = 1; 34 while(!q.empty()) 35 { 36 t = q.top(); 37 q.pop(); //剔除偶数时**。 38 if(ans % 2 == 1) 39 { 40 t.posi+=t.dy; //更新位置,重新入队。 41 q.push(t); 42 } 43 ++ans; 44 } 45 printf("%d\n", t.posi); 46 } 47 return 0; 48 }
标签:
原文地址:http://www.cnblogs.com/fengshun/p/4685455.html