[BZOJ1029] [JSOI2007] Construction Repair

1029: [JSOI2007] Building Repair time limit:4 Sec Memory limit:162 MB
submit:5041 solved:2242
[Submit] [Status] [discuss] Description

Xiao Gang is playing a computer game called "building repair" provided by Jsoi: after a fierce battle, the T tribe wiped out all the Z tribes
Intruders. But there are already n construction facilities in the T Tribe's base that have been severely damaged, and if not repaired soon, the construction facilities will be completely
Destroy. Now the situation is: there is only one repairman at the base of the T tribe, although he can get to any building in an instant, but repairing each building requires
To a certain time. At the same time, the repairman repaired a building to repair the next building, unable to repair multiple buildings at the same time. If a building is in a
The building was scrapped when it was not completely repaired within a period of Time. Your task is to help small just rationalize a repair sequence to repair as much
of Buildings.

Input

The first line is an integer n the next n rows of two integers per line t1,t2 describe a building: repairing the building takes T1 seconds, and if it's within T2 seconds
Without the repair, the building was Scrapped.

Output

Output an integer s, which means that a maximum of S buildings can be Repaired. N < 150,000; T1 < T2 < Maxlongint

Sample Input4
100 200
200 1300
1000 1250
3200Sample Output3This problem uses greedy thought, first for all the number of T2 sort, then maintenance repair of the T1 of the large heap, each looking for the largest heap in the current T1 is less than it, less than replace. Correctness obviously.

1 #include <iostream>2 #include <cstring>3 #include <cstdio>4 #include <cstdlib>5 #include <cmath>6 #include <algorithm>7 #include <queue>8 #include <vector>9 #define LL Long LongTen using namespace std; one int n; a struct Data{int t1,t2;}a[150005]; - BOOL CMP (data c1,data c2){return c1.t2<c2.t2;} - struct CMP1 the{ - bool operator () (data c1,data c2) {return c1.t1<c2.t1;} - }; - priority_queue<data,vector<data>,cmp1> q; + int main () -{ + scanf ("%d", &n); a for (int i=1;i<=n;i++) scanf ("%d%d", &a[i].t1,&a[i].t2); at LL sum=0; - int ans=0; - Sort (a+1,a+n+1,cmp); - for (int I=1;i<=n;i++) -     { - if (a[i].t2>=sum+a[i].t1) {q.push (a[i]); ans++;sum+=a[i].t1;} in Else -         { to if (q.empty ()) continue; + Data now=q.top (); - if (now.t1>a[i].t1) the             { * sum=sum-now.t1+a[i].t1; $ Q.pop ();Panax Notoginseng Q.push (a[i]); -             } the         } +     } a printf ("%d", ans); the }

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[BZOJ1029] [JSOI2007] Construction Repair