CSU 1757 (greedy or tree-like array)

1757: Train inbound time limit:1 Sec Memory limit:128 MB
submit:209 solved:51
[Submit] [Status] [Web Board] Description

Train station people are always waiting for the train to stop at the platform, a platform has a train stay can not have other trains enter, today there are N trains through, known they come in time Si and the outbound time Ti, the station time to the outbound time between the train must have a platform to stop it, ask all trains can stop on time, At least how many platforms to give these trains

Input

The first line enters a positive integer t, which indicates the number of data groups
Enter a positive integer n for the first row of each group of data, indicating the number of trains (n<=10000)
Next n rows, enter 2 positive integers per line si,ti, indicating the stop time and outbound time of the first train (SI<TI<1E9)

Output

The minimum number of stations to be scheduled for each set of data output

Sample Input

131 33 44 6

Sample Output

2


Because up to 20,000 points, the discretization of the tree array, the maintenance interval, the last to find the maximum value of the point.

#include <iostream>#include<cstring>#include<stdio.h>#include<stdlib.h>#include<algorithm>using namespacestd;Const intN =100005;intX[n],a[n],b[n];intC[n],n;intLowbit (inti) {    returni& (-i);}voidUpdateintIdxintv) {     for(inti=idx;i<=2*n;i+=lowbit (i)) {C[i]+=v; }}intGetsum (intidx) {    intsum =0;  for(inti=idx;i>=1; i-=lowbit (i)) {Sum+=C[i]; }    returnsum;}intMain () {inttcase; scanf ("%d",&tcase);  while(tcase--) {memset (c,0,sizeof(c)); scanf ("%d",&N); intCNT =1;  for(intI=1; i<=n;i++) {scanf ("%d%d",&a[i],&B[i]); X[cnt++] =A[i]; X[cnt++] =B[i]; }         intK =2; Sort (x+1, x+CNT);  for(intI=2; i<cnt;i++){            if(x[i]==x[i-1])Continue; X[k++] =X[i]; }         for(intI=1; i<=n;i++){            intL = Lower_bound (x+1, X+k,a[i])-(x); intR = Lower_bound (x+1, X+k,b[i])-(x); Update (L,1); Update (R+1,-1); }        intMAX =-1;  for(intI=1; i<=2*n;i++) {MAX=Max (Max,getsum (i)); } printf ("%d\n", MAX); }    return 0;}

Greedy strategy: For the starting point and the end of the order, meet the starting point of a more platform, meet the end of the word less a platform. The maximum value in the middle procedure is the result.

#include <iostream>#include<cstring>#include<stdio.h>#include<stdlib.h>#include<algorithm>using namespacestd;Const intN =100005;intA[n],b[n],n;intMain () {inttcase; scanf ("%d",&tcase);  while(tcase--) {scanf ("%d",&N);  for(intI=0; i<n; i++) scanf ("%d%d",&a[i],&B[i]); Sort (A,a+N); Sort (b,b+N); intans=0, sum=0; intj=0, i=0;  while(I!=n && j!=N) {if(b[j]<A[i]) {Sum--; J++; }            Else{sum++; I++; } ans=Max (ans,sum); } printf ("%d\n", ans); }    return 0;}

CSU 1757 (greedy or tree-like array)