Ural 1203 Scientific Conference (Greedy | DP)

Scientific Conference

 

I used to brush the computation of ry. I was able to buy the computation of ry even during the invitational competition. During the summer vacation, I went to multiple schools, supplemented the questions of multiple schools, and refreshed the weak DP. If multiple schools have computational ry, they must be eliminated -. -

 

Question: give you the start time and end time of the N reports and ask you how many reports you can listen. The minimum interval between reports is 1.

 

Idea: it is actually a matter of activity arrangement. You can use greedy or DP. It is easier to write greedy. Because DP is practiced, you can write it again using DP.

 

Greedy is a very simple issue of activity selection. Starting from the end time, find the best choice for each activity.

 1 struct node{ 2     int b, e; 3 } N[100005]; 4  5 int cmp(node x, node y){ 6     if(x.e == y.e) 7         return x.b < y.b; 8     return x.e < y.e; 9 }10 11 int n;12 int main()13 {14     scanf("%d", &n);15     for(int i = 0; i < n; ++i){16         scanf("%d%d", &N[i].b, &N[i].e);17     }18     sort(N, N+n, cmp);19     int ans = 0;20     int t = 0;21     for(int i = 0; i < n; ++i)22     {23         if(N[i].b >= t)24         {25             ans++;26             t = N[i].e+1;27         }28     }29     printf("%d\n", ans);30 31     return 0;32 }

Greedy

 

 

DP:

1 struct node {2 int B, E; 3} n [0, 100005]; 4 5 Int CMP (node X, node y) {6 if (X. E = y. e) 7 return X. B <Y. b; 8 return X. E <Y. e; 9} 10 11 int N; 12 int dp [1, 30005]; 13 int K [2, 30005]; 14 int main () 15 {16 scanf ("% d ", & N); 17 int last =-1; 18 for (INT I = 0; I <n; ++ I) {19 scanf ("% d ", & N [I]. b, & N [I]. e); 20 last = max (last, N [I]. e); 21} 22 sort (N, N + N, CMP); 23 for (INT I = 0; I <n; ++ I) 24 {25 DP [n [I]. e] = 1; // the end time of the record is 26 K [n [I]. e] = N [I]. b; // the start time of the activity that records the end time is 27 // There Is A sorting prior to the start time, so the selection will overwrite the optimal 28} 29 for (INT I = 1; I <= last; ++ I) 30 {// DP time 31 if (K [I]) /// if the current time point has ended activity 32 DP [I] = max (DP [I-1], DP [K [I]-1] + 1 ); 33 DP [I] = max (DP [I], DP [I-1]); /// if activity 34} 35 printf ("% d \ n", DP [last]) does not end at the current time point; 36 37 return 0; 38}

DP